Youtube daily report Jan 22 2019

- [Instructor] So in the last video,

we explored the rate law in pretty good detail.

We talked about the importance of the order of reaction

and we talked about the importance of the rate constant.

So let's go ahead and use the same reaction

where we had nitrogen plus three equivalents

of hydrogen becoming two NH threes,

and we said that the rate law for this reaction

is the rate, equals the rate constant

times the concentration

of nitrogen to some power,

we'll call it 'a' here, and the concentration

of hydrogen to a different power, b.

Okay?

Experimentally, we wanna be able to figure out a way

where we can figure out what a and b are,

and then calculate what the rate constant is,

so to have a complete rate law,

we need to know the orders

of the reaction with respect to each reactant

and we also need to know the value

for the rate constant, okay?

And there are a number of ways to do this,

but really kind of like the most universally accepted

and easily accessible is known as the method

of initial rates, okay?

It's known as the method of initial rates,

because we are measuring the rate of a reaction

at the very very beginning, right?

So we mix together a certain amount of our two reactants

and we monitor that rate of the reaction

at the very very beginning,

so it's a place in the reaction

where we have the most control on concentrations, okay?

So what I'm showing here is a table of data, right?

These are three different experiments that we did,

so Experiment One, Experiment Two,

and Experiment Three, and we measured the rate

of the reaction for each of these combinations of nitrogen

and hydrogen, and from this experimental data,

we're gonna be able to determine

the complete rate law including orders with respect

to nitrogen and hydrogen,

and the value of the rate constant, okay?

So the trick to this is,

basically comparing the rate law of two different reactions,

so what we're gonna do is we're gonna take the rate law

for the first reaction,

divided by the rate law for the second reaction, okay?

So in this case, the rate law is,

right, so the rate of the first reaction

is 300, divided by 1200, okay,

so that's rate of the first reaction divided by the rate

of the second reaction.

This equals k, right, so our rate constant,

divided by the rate constant for the second reaction,

which is the same number,

times the concentration of nitrogen for the first reaction

which is three, okay, and this is to the a power,

divided by three to the a power.

Now, the concentration of hydrogen to the b power,

divided by the concentration

of hydrogen to the b power, okay?

So each piece here, right, so the top of this,

we have rate equals k times the concentration

of nitrogen to the a power times the concentration

of hydrogen to the b power.

We divide it by the rate law for the second reaction,

for the second experiment.

And the really nice thing about doing this

is that we can quickly see that this cancels,

the rate constants cancel,

and three to the power of a cancels.

It's the same on top and on the bottom,

so what we're able to do

is to simplify this to 3/12, which is,

so this is 1/4 equals 2/4 or 1/2 to the power of b, okay?

So you may be able to quickly figure out

that, if you take 1/2 squared,

you'll end up with 1/4, but let's go ahead

and go through and mathematically prove this.

So the easiest way to do this

is to take the log of both sides,

okay, so I'm gonna go ahead and move over here,

so I have a little more space.

I'm gonna take a log of both sides.

So the log of 1/4 equals,

when I take a log of this term right here,

this exponent comes out,

so equals b times the log of 1/2, okay?

And then this becomes b equals the log

of 1/4, divided by the log of 1/2.

Plug this into your calculator,

and you will see that b equals two, okay?

So this works really nicely

because we are able to cancel out so many terms

and be left with a mathematical expression

with only one unknown, okay?

Using the rules of logarithmic functions,

we can solve for b and figure out that b is two.

So up here, b is equal to two.

We can do the same thing comparing reactions two

and reactions three, because in this case,

the concentration of hydrogen is the same,

so we're gonna see that that term cancels out

and we'll be able to determine a,

because the concentration

of nitrogen is different in the two reactions.

So let's go ahead and do this, okay?

I'll change colors here, so we're going purple.

So we're gonna do the rate

of the second reaction divided by the rate law

of the third reaction.

So I have 1200 divided by 2400, equals the rate constant

divided by the rate constant,

okay, I have the concentration

of nitrogen to the a power divided by the concentration

of nitrogen to the a power,

times the concentration of hydrogen,

oops, the concentration of hydrogen which is four,

to the b divided by four to the b.

And we see here that our rate constants cancel

and four to the power of b cancels,

giving us 1/2 equals 1/2 to the power of a.

So in this case it's pretty clear that a equals one.

So in our rate law, a equals one.

We now know the order with respect to nitrogen

and the order with respect to hydrogen,

so the last step is to solve for the rate constant,

okay, and this is really straightforward.

Pick any one of the reactions,

either Reaction One, Reaction Two, or Reaction Three.

Pick any of them, and plug the values into the rate law

that includes what we now know is one for the order

of nitrogen and two for the order of hydrogen.

So I'm just gonna pick the first reaction here, okay?

Actually, I'll pick the third one because I can see it.

So our new rate law, rate equals k times the concentration

of nitrogen to the first power,

the concentration of hydrogen to the second power,

okay, my rate is 2400 and be careful with units here now.

So this is molar per minute, right?

Did you notice that up here, the rate is molar per minute?

So this equals the rate constant times the concentration

of nitrogen which is six molar,

this is to the first power,

and the concentration of hydrogen,

which is four molar to the second power, okay?

So we can plug these values into our calculator,

we have 2400 divided by six divided by four squared,

and, oops,

and we're gonna see that our rate constant

is equal to 25 and now,

if we're careful with units, what you'll see

is that the rate constant ends up having the units of molar

to the negative two minutes to the negative one,

and this equals one over molar squared times

one over a minute, okay?

So the rate constant is 25 per molar squared, per minute.

For more infomation >> Method of Initial Rates - Duration: 9:09.

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Integrated Rate Laws and Plotting Data - Duration: 8:53.

- [Instructor] Okay, so in the previous video we saw that

each reaction order has it's own unique integrated rate law,

which is really useful

because from the integrated rate law,

you can determine if a reaction is zeroth order,

first order, or second order through experimental data.

Okay?

And what I'm showing here is just a quick summary

of these rate laws and integrated rate laws

for the different reaction orders.

Note that all of the integrated rate laws adopt a form of

y equals m X plus b,

and the slope of all of the integrated rate laws

are related to the rate constant.

So for zeroth order and first order,

the slope of the line that comes out when plotting

concentration or natural log of concentration versus time

is the negative of the rate constant.

For a second order reaction,

if you plot one divided by the concentration versus time,

the slope is a positive value and that is the rate constant.

So what we're gonna do in this video

is take some experimental data

and use these equations to figure out if a reaction is

zeroth order, first order, or second order,

and then also determine the rate constant.

So I'm going to switch in over to Excel here,

and once we're done with Excel, I'll also show you how

to do this in Google Sheets,

at least the graphing part of it,

because it's little bit different than this.

Okay, so what I have here is some experimental data

for the decomposition of N2O5.

'Kay?

So we have the time here,

which we're gonna plot on the x-axis,

and the concentration of N2O5, which is our reactant,

as time proceeds.

And we notice this concentration is decreasing,

as you would expect,

because N2O5 is being consumed when it decomposes.

So if this is a zeroth order reaction,

we can plot this time versus concentration

and it should be a straight line if it's zeroth order.

So let's try this.

So I selected the data here,

and then if you're in Excel

go ahead and go up to Insert,

and under Recommended Charts, I like to do the Scatter.

So pick this Scatter right here,

and what we see is that this line

is absolutely curved and not linear, right?

You can see that curvature.

So this is not a zeroth order reaction

because this line is not linear.

So I'm gonna delete this graph

because it's not zeroth order.

Okay, let's see if it's first order.

So I'm gonna copy the time data,

so I'm hitting Control-C here,

and copying that data and Control-V, that's copy and paste.

Mmkay?

And then in this column I want the natural log

of the concentration of N2O5.

Okay, so this is where Excel is really great

because you can set up an equation

for Excel to do this math for you.

You click the equals sign,

and that signals Excel this is an equation,

so we wanna do the natural log, so equal L-N,

and then open parentheses,

and then I'm gonna select the cell

that contains the concentration of N2O5.

And in my spreadsheet, that's cell E7.

And when I close that parentheses and click Enter,

then it calculates the natural log of 0.1 for me.

So the natural log of 0.1 is -2.306.

Okay?

I wanna do that same thing for all of the other time points.

But another really good benefit of Excel

is that I don't have to manually do it.

I'm gonna select this cell,

and I'm gonna put my cursor

over the bottom-right corner of this box,

and you'll see that cursor changes into the crosshair.

So when it's a crosshairs, click and drag it down.

And that will do the same calculation

for all of the other times.

And now we can plot this as a scatter plot,

so select the data, click Insert, and plot a Scatter.

And what we see here is, this does look pretty darn linear.

So it's pretty likely that this is gonna end up being

a first order reaction.

Mmkay?

So we're gonna keep this plot up.

And just to make sure,

we're gonna plot and see if it's second order.

So for second order, I'm gonna copy the time data,

and then I wanna have

one divided by the concentration of N2O5.

Okay, so I need to calculate this,

so I'm gonna click equal, which signals Excel

that this is gonna be an equation,

and I wanna do one divided by the concentration of N2O5,

which is cell E7 for me.

When I do that, it takes one divided by 0.1,

which has a value of 10.

Now I'm gonna do the same thing I did with natural log,

I'm gonna select that cell.

In the bottom-right corner I'm gonna click and drag it down,

and that's gonna update all of the other cells

so that they do one divided by concentration as well.

Let's plot it.

So we're gonna select this data,

Insert,

and select a Scatter plot.

Mmkay? So as you can see here,

we definitely have curvature in this line,

so this is not linear so it's not second order.

So indeed, we have determined that this reaction,

the decomposition of N2O5, is first order

because the natural log of concentration versus time

is linear.

So the last step would be to determine what the slope

of this line is so that we can determine the rate constant.

Okay?

So I'm gonna right-click,

or if you're a Mac user Control-click,

on one of the points,

and you'll have this Add Trendline option.

Okay, when you do this, if you scroll down

and select the Display Equation on Chart option here,

okay, so that's telling the software

to go ahead and put that equation right on your graph,

which we've done here.

Okay?

So the slope of this line is -0.0069,

and the y-intercept is -2.3026.

So we remember from our integrated rate laws, right,

that for a first-order reaction the slope is -k,

and our slope was -0.0069,

so the rate constant for this reaction is 0.0069,

and the unit would be one divided by second

because the time is in seconds

and a first-order rate constant has the units

of one divided by time.

Okay?

So let's see how you can plot this in Google Sheets

if you don't have access to Microsoft Excel.

So what I'm gonna do here is I'm copying this data,

equations and all, 'cause you would do up to this point

exactly the same way as we did in Excel.

But I'm gonna put it into a Google Sheets spreadsheet.

Okay?

So here we're looking at Google Sheets, 'kay?

I've done everything the same.

What I wanna do here is plot each of these,

but in this case I know

that I only need to plot the first order.

So I'm gonna select my data,

choose Insert,

Chart.

'Kay, so I wanna change the chart type to a Line.

Okay?

And the last thing you wanna do,

alright, so here you can very clearly see

this a straight line,

and the last thing you wanna do

to get the y equals M X plus B equation,

is go to Customize,

Series,

okay, you wanna click Trendline,

and then scroll down and have the Label use Equation.

When you do that, you see this new box pop up here,

and this tells you

y equals -6.92 times 10 to the minus third X,

plus -2.31,

so the slope is -6.92 times 10 to the minus third,

so the rate constant would be

6.92 times 10 to the minus third.

For more infomation >> Integrated Rate Laws and Plotting Data - Duration: 8:53.

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Rate Laws and Arrhenius Review - Duration: 8:10.

- [Instructor] In the last video

we introduced the idea of rate laws.

Lets explore the synthesis of amonia as an example here.

Okay, so, when nitrogen and hydrogen react together

we make ammonia gas.

So, if I were to ask you to write a rate law

for this reaction, the way that you would have to start

is by putting rate equals K.

So, the rate equals K, and remember, K is our rate constant.

This is always true of every single rate law.

So, K is our rate constant.

We'll talk a little bit more about this in a second here.

To finish up our rate law, we would then take

the two reactants and take the concentration

of one reactant to a certain power, here I'll put A

times the concentration of the other reactant

to a different power, I'm putting this B here.

And this is the form that every single rate law will adopt.

Rate equals the rate constant times the concentration

of reactants to some power.

These exponents, A and B

are known as orders.

So, if

we have one, right, if A equals one,

then it would be true that this would be first order

with respect to nitrogen.

If B were equal to two, this would be second order

with respect to hydrogen.

And the overall order of the reaction is the sum.

The overall order

would be one plus two,

so my overall order would be third order.

So this reaction would overall be third order.

It would be first order with respect to nitrogen,

second order with respect to hydrogen.

We can rewrite our rate law where rate equal K

times the concentration of nitrogen to the first power.

The concentration of hydrogen to the second power.

This would be a rate law.

Okay, so, let's think about what these orders mean.

And if we look at this equation,

hopefully you can draw the conclusion

that the higher the order, so, large orders

means that

there is a lot

of influence

on reaction rate.

So, if something has a large order,

then it means that the rate of the reaction

is very sensitive to changing concentration.

So, in this case, we have first order

with respect to nitrogen, second order

with respect to hydrogen.

So, the rate of the reaction is more sensitive

to changing the concentration of hydrogen.

It's less sensitive to changing

the concentration of nitrogen.

That's the idea of order.

The higher the order, the more influence

concentration has on rate.

If you have a really high concentration,

you'll have a really high rate.

But as the concentration decreases,

the rate decreases much more quickly for second order

than it does for first order.

The rate constant is a really important idea.

To explore the idea of the rate constant

we're gonna go back to the reaction coordinate

that we introduced in the last grouping of videos.

Remember in our reaction coordinate we were plotting

energy versus reaction progress.

And we can start with our reactants.

We end with our products, and we said

we transitioned through this transition state.

Our path goes up through the transition state

before we drop back down to make our products.

We have to go through a high-energy state.

And this high-energy state, the energy that it takes

to get there is called the activation energy.

Which remember, we gave a capital E, lower case a.

So, the activation energy is E sub a.

The rate constant is very much related

to this activation energy.

And I'm just gonna show an equation here.

For this class, you won't have to use the equation

but I'd like you to see this relationship.

The rate constant is equal to this parameter here.

This is called the eranious property.

This is called the eranious parameter.

And don't worry about it.

But it's E to the minus E A

over RT.

So, the important thing to note here

is that the rate constant is sensitive

to the activation energy and the temperature.

The idea is that as you increase the activation energy,

the rate constant decreases.

This whole exponent

becomes more negative

as the activation energy becomes larger.

And the more negative the exponent,

the smaller the rate constant.

So, for a large activation energy,

we have a really small rate constant.

But if we have a really small activation energy,

we have a really large rate constant.

The rate constant increases as activation energy decreases.

The opposite's true for the relationship

between temperature and the rate constant.

Temperature is on the bottom here.

Let's go ahead and kinda clear this up a little bit.

Temperature is on the bottom here.

The importance of temperature

is that the higher the temperature is,

the more likely it is that individual molecules

that react together will have enough energy

to get up over this activation hump.

The higher the temperature, the more likely

it is the molecules will have enough energy.

So, a big temperature means a big rate constant.

A small temperature means a small rate constant.

So, the rate constant of the reaction

is sensitive to the size of activation energy

and the temperature of the reaction.

Decreasing the activation energy

will increase the rate constant

which will in turn increase the rate of the reaction.

If we increase the temperature,

that leads to an increased rate constant,

and that leads to an increased rate.

Those are the only two ways to change the rate constant:

Changing the temperature or changing the activation energy.

Other than that, every single reaction, right,

so each reaction has its own unique rate constant,

and the only way to change it would be

to change the temperature or to somehow

influence the activation energy.

In the next video we are going to explore ways

where we can actually determine the rate law

for a reaction, right, so,

we need to be able to figure out what these orders are,

and we need to be able to calculate a rate constant

so that we can use the rate law

to determine the rate of a reaction.

For more infomation >> Rate Laws and Arrhenius Review - Duration: 8:10.

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Integrated Rate Laws - Duration: 8:13.

- [Instructor] In the previous set of videos,

we explored the rate law.

We discussed how a rate law takes the generic form,

rate equals a rate constant times the concentration

of some reactant to some power,

and then you could have other reactants

to different powers as well.

So, this rate law is really, really useful

if we want to determine the rate of a reaction

at a given concentration of our reactants.

But it's not very useful if our goal is

to figure out how much reactant remains

after the reaction has preceded

at a certain amount of time.

To do that, what we want to do is

access our integrated rate laws.

So, for those of you who have

a little bit of calculus background,

you might recognize this word "integrate"

as something that you learn when you learn

how to do calculus.

In this class, we're not gonna do the calculus,

we're just gonna see the result of

integrating this rate law.

There are three forms of integrated rate laws

that you need to be familiar with.

So, if a reaction is zeroth order,

then the integrated rate law is

the concentration of A equals minus kt

plus the initial concentration of A.

This would be the concentration of the reactant

at any specific time,

and that time is t.

This term right here is the initial concentration.

So, this is the concentration at time zero,

and k is the rate constant.

So that's the zeroth order.

If we have a first order reaction,

the integrated rate law takes a similar form,

but in this case it's the natural

log of the concentration equals minus kt

plus the natural log of that initial concentration.

All of the same terms here,

concentration, rate constant, time, and

initial concentration, but now we have

the incorporation of a natural log.

And for a second order reaction,

we end up with one divided by

the concentration of A, equals positive kt

plus one divided by the initial concentration.

These are our integrated rate laws,

and they're really useful, because

if you know the rate constant,

and you know a specific time

and you know what the initial concentration

of your reactant is,

then you're actually able to figure out

the concentration of the reactant

after any amount of time has passed.

It becomes just a plug and chug type of problem,

here is the equation, use it.

But the other really, really useful form of this

I guess, the other really useful application,

is that

it can be used to determine

what the order of the reaction is

with respect to one reactant.

Because what you may notice here

is that each of these adopt a form of

y equals mx plus b.

This is the equation for a straight line.

If a reaction is zeroth order

with respect to the reactant A,

it will obey the zeroth order integrated rate law.

And what this means is that

if we were to plot on an x y coordinate

the concentration of the reactant versus time,

we would see a straight line.

If it's zeroth order, we would see a straight line

and this line would have the slope

of negative k.

You would have the y-intercept of

the initial concentration.

From our equation up here,

we see the concentration of A,

which is on our y axis,

equals minus k, which is our slope,

times t, which is our x,

plus this initial concentration of A,

which is our y-intercept.

So the y-intercept is the

initial concentration of A,

the concentration at times zero,

we plot time on the x axis,

the slope of the line is minus k,

so from this, monitoring the concentration

of a reactant versus time,

if it's zeroth order, it will be

a perfectly straight line,

and the slope of that line will be

the negative of the rate constant.

If it's not a zeroth order reaction,

let's say it's actually a first order reaction.

What you would see is a non straight line.

If it's a second order reaction,

you would see a line that's also not straight.

The only way a concentration of A

versus time graph is gonna be linear

is if the reaction is zeroth order

with respect to that reactant.

We can apply a similar logic to thinking about

plotting the natural log of

a concentration versus time.

So, on this graph, we'll say we have

the natural log of the concentration of A versus time.

Now, in this case, natural log of A is our y,

time is still our x, the slope, we're gonna see,

is still negative k,

and our y-intercept will be

the natural log of A.

So, if we were to plot this,

it would be a perfectly straight line now.

The slope of this perfectly straight line

would be minus k, so that would be our slope,

and the y-intercept here would be

the natural log of our initial concentration.

If the reaction were zeroth order,

it would not be a straight line,

it would be curved again.

If it were second order, again,

it would not be a straight line,

it would be curved.

So, the only way this would a straight line

is if it's first order.

Let's think about second order.

Just like we saw before,

if we plot

one divided by concentration of A versus time,

then the slope of this line should be

now positive k,

and one over the initial concentration

should be our y-intercept.

Plotting one over the concentration of A versus time,

if it's a second order reaction,

then this will be a straight line.

And notice now the slope is positive,

because the slope is the rate constant,

and rate constants are positive values.

The y-intercept here would be one over

the initial concentration.

So, these integrated rate laws are really useful

because they have a direct linear relationship

between the concentration of reactants and the time,

allowing you to figure out

exactly what the concentration of a reactant is

at any time, if you know the rate constant.

They're also useful, because

if you don't know the order, and you plot

concentration of A versus time,

or natural log of concentration of A versus time,

or one divided by concentration of A versus time,

one of them will be linear,

and then you can conclude that the

order of that reactant is zeroth, first, or second.

For more infomation >> Integrated Rate Laws - Duration: 8:13.

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#WhatsOnFios with Kristian and Mark Ep. 49 | Series Premieres & WWE Royal Rumble - Duration: 3:09.

Hello everyone, and welcome back to #WhatsOnFios.

That is Kristian, I am Mark.

This vid is sponsored by Verizon Fios.

We like to take you through all the latest movies and TV offerings, and Kristian, there's

a lot of TV shows that are coming back for new seasons.

A pair of season threes as a matter of fact.

First one I'm excited to talk about, I know you are as well.

Victoria, on PBS.

I wouldn't have watched this had it not been for my wife, and I'm so glad that I did.

The first two seasons really engaged.

Cannot wait for season three.

Yeah, Jenna Coleman plays Victoria.

Any time you have a lot of political skirmishes, you mix that with some royal intrigue, conspiracies,

it just makes for good television.

Something that you can definitely catch on HBO, is gonna be True Detective.

Now, True Detective, I mean this thing came out with such a bang when season 1 hit, and

now you have Mahershala Ali stepping in.

He's already got a shelf full of awards.

I'm looking forward to it a lot.

Oh man, I was on board when you told me True Detective season 3 was coming out but then

Mahershala Ali's gonna be -

Stephen Dorff too, plays his partner.

I like Stephen Dorff.

Mahershala Ali in the lead here and already of

just the trailers that I have seen, it is

one of my most anticipated things, not just television, not just film, it's just one of

my most anticipated things this season.

Yeah, powerful, tense, and you talk about season threes for these two shows, how about

season nine for Showtime's Shameless.

You've been on the Shameless train since day one.

William H. Macy and Emmy Rossum.

They're back for more carnage.

Yeah, it's family, isn't it?

It is destructive family, and this, once again all the seasons that they've been together,

what else can happen?

The fact that they kept you really locked in for all these seasons is a testament to

how good the show has been.

It's kinda like if you and I had to live - have we ever lived together?

We've never been roommates, have we?

No, I would never sign on to that.

Alright, well thanks for endorsing my roommateship.

There's more than just television on this month, my man.

What do you got? January.

WWE's Royal Rumble is gonna be on.

I am a big wrestling fan.

I like the WrestleManias, I like Survivor series, but the Royal Rumble, that's my jam.

That's the one I look forward to.

Whoever wins gets a shot at main event at WrestleMania.

He's very excited about it.

It's Sunday, January 27th, and I'm gonna surprise you with a couple of quick trivia questions.

You ready?

Trivia questions?

Are these gonna be tough ones?

I studied these and you have to answer them.

Who is the former GM of Raw, who is now competing in the Royal Rumble?

That is Baron Corbin.

You're one for one.

Who plays guitar, who's competing in the -

I would like some hard ones sir.

That would be Elias.

Okay, how many Royal Rumbles has John Cena won?

Ooh, two.

He has won two.

Hey, well done buddy.

I knew that one.

Do you have any trivia for me?

When can you watch the Royal Rumble?

You can watch the Royal Rumble Sunday, January 27th.

That was a great set up.

On pay-per-view, so make sure you catch that.

As far as the TV shows go, comment below.

Let us know, what are you looking forward to.

What are the new seasons, and you can catch past episodes on demand and like we said,

the WWE Royal Rumble is Sunday, January 27th on pay-per-view.

Thank you all for tuning in to this episode of #WhatsOnFios.

Make sure you come back for all the latest in blockbuster movies, popular TV, and more.

You can check out the link in this video's description for all the latest offerings and

information from our friends at Fios.

For more infomation >> #WhatsOnFios with Kristian and Mark Ep. 49 | Series Premieres & WWE Royal Rumble - Duration: 3:09.

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Destruction Room offering stress relief for furloughed workers - Duration: 1:43.

For more infomation >> Destruction Room offering stress relief for furloughed workers - Duration: 1:43.

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EPIC treasures around a BIG rock! Metal detecting Finland - Duration: 9:02.

If I would be burying something here on this field, I would choose that rock

Let's go and check it out. Probably filled with treasures underneath it

Let's go. first signal one of these little pretty buttons

Good start. Really steady signal of

16-17

Let's dig it up see what it is.

I of course place my bet that it's some kind of crap.

Because I only dig up crap.

And

A bottle cap

Second in a row.

pretty good signal

Let's see, maybe this will be my treasure

Interesting, look we have something we have roundness

We have roundness, our first coin!

It's a beautiful beautiful coin

Alexander ii

It's 5 Penniä (Finnish Penny). A beautiful 5 penniä. here I will put

some water on it check it out

Cleaned it up a little bit. You still can't see the year

so

But it's Alexander ii

Who was the russian tsar.

when

finland belonged to

Russia. And it should be from 18,

Was it 65? until

1875 something like that. So I think I can see a 7

18

so it has to be

1867

5 Penniä. These are nice coins always

Happy to find these in

Finland ah

Pretty coin

you can see the big A

that stands for

Alexander and then the

Two in Roman letters

Nice one.

One of the lowest lowest signals I have had

today but had to dig it up and I can see some roundness here.

But I don't think it's a coin it's maybe a button. It's a button.

Not that old. One of these

usual buttons that you find maybe I would guess fifty years sixty years

My second button.

Well, this doesn't look old at all.

Some good signals here a lot of iron I bet I will find something cool. Woo

my first steady signal in a while and

These are always nice to find

this is a

lead

bag seal.

They put some grain in the bags and put a seal on it

This has S

S-something and a M. You could probably know where it has come from if you read this always nice finds

Trying to search

Around this big old rock here

Have one really shit signal, but you know

Have to check. Could be a treasure chest then I just broke it

Not foil.

But a

Piece of iron.

Hmm

New interesting signal. Right next to the rock

Maybe this will be my treasure

Treasure

I can feel it.

I see something. Yep. I could see something, and that's a rusty nail

My father found one coin next to the big rock

In

bad shape

it looked like a small Roman coin at first, but I think it's a one penny from

end of 1800s probably

nice

Have a

interesting object in there

Could be, it's a big iron object could be an axe

An axe or something?

It's really deep. I have to dig it up. Oh, it was so deep, but I got it out.

That was a work out

I

already

imagined the

sword or an axe but

it was just, I think it's like a

I don't know the English name. You sharpen something with it

Cool find anyway. Well now now I have the best signal for today, steady

34 could be a coin it's somewhere on the surface should be close to the surface

It could be some kind of trash

Deeper

Yeah copper

trash

no wonder it gave such a good signal.

Thought I had something interesting but no no, no just yeah

What would this be? Piece of some farming equipment again? I have picked up so much

Farming equipment soon, that I can soon become a farmer instead.

I will skip this treasure hunting, build my own

Farming machines and

Buy a farm and yeah, maybe a cow or two, some sheeps and

Grow my own food

Mmm, I think I see a coin

Think I see a coin over here

Yes, it is

Yes, it is

What coin?

I have no idea. Really bad

condition probably one of the one pennies in the end of 1800's

Finnish pennies, my second coin pretty good

found this interesting piece of

Something no idea what it is again, but I think

This is an old

golf

pin, you know

when you start your rounds you put this one in the ground. Put the gold ball here and

Swoosh

Hole in one

Yeah, you know

hundred years ago.

A weird signal

Dug up this. This is a bottle cap? Yeah, I think it's a bottle cap.

It looked like a silver coin when I saw it like this

This might be one of my last signals

We have it pinpointed.

And foil

Okay, that might be my last

signal for today before it gets dark. I wanna go to the sauna. See you in the next video

remember to subscribe, like, share, comment and

See you in the next video

For more infomation >> EPIC treasures around a BIG rock! Metal detecting Finland - Duration: 9:02.

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Alabama - 40-Hour Week - guitar cover (кавер на гитаре) - Duration: 4:31.

And hello again.

My name is Sergey,

I play acoustic guitar covers of classic rock songs.

Country group Alabama.

I'll be brief.

Today, a cover melody called 40-Hour Week.

This song was composed in 1985.

She climbed high enough in the charts.

Dedicated to this song blue collar.

This song ends with a quote,

a quote from the very famous American melody America Is Beautiful.

So, the cover of the band Alabama sounds.

Song 40-Hour Week.

For more infomation >> Alabama - 40-Hour Week - guitar cover (кавер на гитаре) - Duration: 4:31.

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Intro to Making Money on YouTube - Duration: 3:03.

Hi, my name is Stefan, and I'm a Monetization Strategist here at YouTube.

The question I often hear from creators is "when can I start making money on YouTube?"

So I'm going to explain the basics of what you need to know

to start earning money with your YouTube channel.

Before you can earn money on your channel, you need an audience.

Now, to build an audience

you need to make consistent videos that viewers really like watching.

And, like anything on YouTube

it's important that your videos follow YouTube's Community Guidelines.

After that's all working for you

you can apply to join the YouTube Partner Program, or YPP

which lets creators monetize their videos on YouTube.

To qualify for YPP, you need to be in good standing with YouTube

and have 4,000 valid watch hours in the previous 12 months

and at least 1,000 subscribers.

Now, we've set these thresholds

because we want creators to be good citizens on the platform

and these requirements

ensure our teams have enough information to really review your channel.

Now, you can apply to YPP at any time, but once you hit these stats

your channel will be reviewed against our policies

and if everything looks good, you can start earning money from ads

and from YouTube Premium subscribers watching your content.

Also, once you're approved for the YouTube Partner Program

there are other revenue streams besides ads that you can tap into.

Things like Channel Memberships or Super Chat.

As long as you meet their criteria and they're available in your country.

So that's the program in a nutshell.

Now I'm gonna walk you through some of the most common questions we get.

The first is around channel reviews.

Say you think your channel qualifies for YPP but it hasn't been reviewed yet.

Well, we usually get back to you in about a month

once you hit the review thresholds

but you can find more about your status on the monetization page in YouTube Studio.

Until then, keep making great content and focusing on building your audience.

Another question that comes up, what if your channel isn't approved?

If you're not approved for YPP

it's likely because your channel doesn't meet

the YouTube Partner Program policies and Community Guidelines.

In which case, you can re-apply again in 30 days.

Before you re-apply, we suggest you read through these policies.

There are some links in the video description.

And go to the monetization page in your account and review your videos

with our YouTube Partner Program policies and Community Guidelines in mind.

The next step is to edit or delete any videos that violate these policies.

Once you're successfully in the program, creators often ask

"Do I have to meet this threshold every year?"

The short answer is, no.

However, you do need to stay active with uploads and community posts.

If you're inactive on your channel for 6 months and fall below the threshold

we may remove your channel from the Partner Program.

So, are you ready? Apply to join today

and stay tuned to this channel and the Creator Academy for more insights

including how to get the most out of ads and other monetization features.

Thanks, and bye for now.

For more infomation >> Intro to Making Money on YouTube - Duration: 3:03.

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Stunning Beautiful Manufactured Home Gold Star 2460 203 with 3 Bedrooms from Champion Homes - Duration: 3:23.

Stunning Beautiful Manufactured Home Gold Star 2460 203 with 3 Bedrooms from Champion Homes