How do I calculate the amount from the volume?

[Math_312 46]

The gold pavilion in the city of Kyoto is covered with gold leaf. Gold leaf has a thickness of 100 nm (nanometres) and is made from gold bars. A gold bar has a volume of 0.65 dm^3. How many square metres can the gold leaf from a gold bar cover?

 

[topsquark]

The gold leaf is (presumably) a rectangular solid, ie, flat. So V = Ah.

-Dan

 

[Opalg]

The gold pavilion in the city of Kyoto is covered with gold leaf. Gold leaf has a thickness of 100 nm (nanometres) and is made from gold bars. A gold bar has a volume of 0.65 dm^3. How many square metres can the gold leaf from a gold bar cover?

Your first task should be to express everything in the same units. So how many nm are there in a dm?

Next step: how many nm^3 in a dm^3? (You'll be getting some large powers of 10 at this stage.)

If the volume of a gold bar is X nm^3 then the formula V=Ah (with H = 100nm) tells you that the area covered by the gold leaf will be given by 100A = X.

Finally, that area A is in units of nm^2. You need to convert that to m^2 to get the result in square metres.

 

[Math_312 46]

Your first task should be to express everything in the same units. So how many nm are there in a dm?

Next step: how many nm^3 in a dm^3? (You'll be getting some large powers of 10 at this stage.)

If the volume of a gold bar is X nm^3 then the formula V=Ah (with H = 100nm) tells you that the area covered by the gold leaf will be given by 100A = X.

Finally, that area A is in units of nm^2. You need to convert that to m^2 to get the result in square metres.

Is the answer 0,01m^2?
If this is wrong can you help me?

 

[Opalg]

Is the answer 0,01m^2?

That really doesn't seem very likely, does it? In fact, 0.01m^2 is one hundredth of a square metre. That is less than the size of a sheet of paper. If you take a gold bar and hammer it out until it is microscopically thin (which is what gold leaf is), you would expect it to cover quite a large area, certainly many hundreds of square metres.

If this is wrong can you help me?

A decimetre (dm) is one tenth of a metre. A nanometre (nm) is one billionth of a metre. So there are $10^9$ nm in a metre, and therefore $10^8$ nm in a dm. When you cube that in order to get a measure of volume, you find that there are $10^{24}$ nm^3 in a dm^3. So a volume of $0.65$ dm^3 is equal to $0.65\times10^{24}$ (or equivalently $6.5\times10^{23}$) nm^3.

Can you continue from there?

 

[Math_312 46]

That really doesn't seem very likely, does it? In fact, 0.01m^2 is one hundredth of a square metre. That is less than the size of a sheet of paper. If you take a gold bar and hammer it out until it is microscopically thin (which is what gold leaf is), you would expect it to cover quite a large area, certainly many hundreds of square metres.

A decimetre (dm) is one tenth of a metre. A nanometre (nm) is one billionth of a metre. So there are $10^9$ nm in a metre, and therefore $10^8$ nm in a dm. When you cube that in order to get a measure of volume, you find that there are $10^{24}$ nm^3 in a dm^3. So a volume of $0.65$ dm^3 is equal to $0.65\times10^{24}$ (or equivalently $6.5\times10^{23}$) nm^3.

Can you continue from there?

Sorry in advance but I find maths very difficult. But according to my calculations based on your explanation, it came out like this:

6.5 x 10^23 / 100 = 65 e 21 (65 and 21 zeros)

And thus I get the answer that 65 e 21nm² can the gold leaf of a gold bar cover?

 

[Opalg]

Sorry in advance but I find maths very difficult. But according to my calculations based on your explanation, it came out like this:

6.5 x 10^23 / 100 = 65 e 21 (65 and 21 zeros)

And thus I get the answer that 65 e 21nm² can the gold leaf of a gold bar cover?

You left out the decimal point, so it should be 6.5 e 21 nm^2. But the question asks for the answer in square metres, so you still have to divide by $\bigl(10^9\bigr)^2 = 10^{18}$ to get the answer in those units.

 

[Math_312 46]

You left out the decimal point, so it should be 6.5 e 21 nm^2. But the question asks for the answer in square metres, so you still have to divide by $\bigl(10^9\bigr)^2 = 10^{18}$ to get the answer in those units.

Ah, I see. so if I convert 6.5 e 21 nm² to m², which then becomes 65000 m². Is this the answer or is something wrong?

 

[Opalg]

Ah, I see. so if I convert 6.5 e 21 nm² to m², which then becomes 65000 m². Is this the answer or is something wrong?

Nearly there! You have one too many zeros. It should be 6.5 e 3 = 6500 m^2.

 

[Math_312 46]

Nearly there! You have one too many zeroes. It should be 6.5 e 3 = 6500 m^2.

So to be sure the answer is 6500 m^2 of the whole question?

 

[Opalg]

So to be sure the answer is 6500 m^2 of the whole question?

Yes. :)

 

[Math_312 46]

Yes. :)

Thank you so much, you are a lifesaver:love:

I don't want to sound annoying, but when I was trying to solve another difficult task, I was stuck on how to figure out the lineup itself.

10.2 n + 10.8 2n / n + 2n

Question: In a survey, several measurements were taken and the average of the measurements was 10.2. In a new survey, twice as many measurements were taken and the mean of these measurements was 10.8. What is the mean of all the measurements?

It's all right if you don't want to help me with another question.