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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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How many grams are there in 3.4 x 10^24 atoms of He?
Lelechka [254]

Given :

Number of He atoms, n = 3.4\times 10^{24} atoms.

To Find :

How many grams are their in given number of He atoms.

Solution :

We know, molecular mass of He is 4 g. It means that their are 6.022 \times 10^{23} atoms in 4 g of He.

Let, number of gram He in 3.4\times 10^{24} atoms is x , so :

x = \dfrac{3.4\times 10^{24}}{6.022\times 10^{23}}\times 4\\\\x =  $$22.58\ g

Therefore, grams of He atoms is 22.58 g .

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An industrial manufacturer wants to convert 175 kg of methane into HCN. Calculate the masses of ammonia and molecular oxygen req
Tema [17]
Given:

175 kilograms of Methane (CH4) to be synthesized into Hydrogen Cyanide (HCN)

The balanced chemical equation is shown below:

2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To calculate for the masses of ammonia and oxygen needed, our basis will be 175 kg CH4.

Molar mass:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol

mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol 
mass of NH3 = 185.94 kg NH3 needed

mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
mass of O2 = 525 kg

mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
mass of O = 131.25 kg O 
4 0
3 years ago
Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2
lapo4ka [179]

Answer: Concentration of CO = 0.328 M

Concentration of Cl_2 = 0.328 M

Concentration of COCl_2 = 0.532 M

Explanation:

Moles of  CO and Cl_2 = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of CO and Cl_2  =\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M

The given balanced equilibrium reaction is,

                            CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.          0.860M     0.860M           0

At eqm. conc.    (0.860-x) M  (0.860-x) M     (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get :

4.95=\frac{x}{(0.860-x)^2}

By solving the term 'x', we get :

x =  0.532 M

Thus, the concentrations of CO,Cl_2\text{ and }COCl_2 at equilibrium are :

Concentration of CO = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of Cl_2 =  (0.860-x) M  =  (0.860-0.532) M = 0.328 M

Concentration of COCl_2 = x M = 0.532 M

3 0
3 years ago
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