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Anvisha [2.4K]
3 years ago
7

How many atoms of gold (Au) are equal to 212g of gold?

Chemistry
1 answer:
Lostsunrise [7]3 years ago
8 0

Given :

Mass of gold ( Au ) is 212 gram.

To Find :

Atoms of gold present in 212 gram of gold.

Solution :

Molecular mass of gold is 197 gram.

So, number of moles of gold in 212 gram is :

n = \dfrac{212}{197} \ moles\\\\n = 1.076 \ moles

Now, we know 1 mole of any element contains 6.022 \times 10^{23} atoms.

So, number of atoms present in 1.076 moles are :

N = 1.076 \times 6.022\times 10^{23}\\\\N = 6.48 \times 10^{23}

Hence, this is the required solution.

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Co diffuses 1.25 times the rate of diffusion of c02
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An oxide of phosphorus contains 56.4% phosphorus and 43.6% oxygen. It's relative molecular mass is 220. Find both the empirical
sladkih [1.3K]
Moles of P = 56,4g/30,974g/mole = 1,82 moles P
moles of O = 43,6/15,999 = 2,73 moles of O

converting to the simplest ratio:
For P : 1,82/1,82 = 1
For O : 2,73/1,82 = 1,5

1 P and 2 oxygens.
PO2 -> the empirical formula

hope this help
7 0
3 years ago
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What is the definition of physical science?
katovenus [111]

Answer:

The definition of physical science is the sciences concerned with the study of inanimate natural objects, including physics, chemistry, astronomy, and related subjects.

Explanation:

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What is the gram formula of na2co3
makkiz [27]

Answer:

106 gfm

Explanation:

element. number masses

Na. 2 23×2

C. 1. 12

O. 3. 16×3

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106 gfm

5 0
2 years ago
For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0
3 years ago
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