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Aneli [31]
4 years ago
15

Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.

54 cm. (1 f-12 inches)
Chemistry
1 answer:
Jlenok [28]4 years ago
6 0

<u>Answer:</u> The density of gold in g/cm^3 is 19.22g/cm^3

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}

We are given:

Density of gold = 1200lb/ft^3

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into g/cm^3, we get:

\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3

Hence, the density of gold in g/cm^3 is 19.22g/cm^3

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<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

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For the given chemical reaction:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

The expression of K_c for above reaction follows:

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}         ........(1)

We are given:

[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M

[CO_2]_{eq}=0.40M

Putting values in above equation, we get:

K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

                      CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial:              0.1      0.1                 0.4       0.1

At eqllm:       0.1+x   0.1+x           0.4-x      0.25-x

Putting values in expression 1, we get:

4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

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