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3 years ago

A gas has a volume of 45 L at a pressure of 140 kPa. What is the volume when

Chemistry

1 answer:

suter [353]3 years ago

8 0

Answer:

21 L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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A 2.832 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 6.439 grams of CO2 and 2

strojnjashka [21]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_{12}O_2$ respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=6.439g$

Mass of $H_2O=2.636g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 6.439 g of carbon dioxide, $\frac{12}{44}\times 6.439=1.756g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.636 g of water, $\frac{2}{18}\times 2.636=0.293g$ of hydrogen will be contained.

Mass of oxygen in the compound = (2.832) - (1.756 + 0.293) = 0.783 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.756g}{12g/mole}=0.146moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.293g}{1g/mole}=0.293moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.783g}{16g/mole}=0.049moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.049 moles.

For Carbon = $\frac{0.146}{0.049}=2.97\approx 3$

For Hydrogen = $\frac{0.293}{0.049}=5.97\approx 6$

For Oxygen = $\frac{0.049}{0.049}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 116.2 g/mol

Mass of empirical formula = 58 g/mol

Putting values in above equation, we get:

$n=\frac{116.2g/mol}{58g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(6\times 2)}O_{(1\times 2)}=C_6H_{12}O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_6O$ and $C_6H_{12}O_2$ respectively.

7 0

3 years ago

What is the diet of sloth bears ?

Svetllana [295]

This is what I got hope it helps

4 0

3 years ago

Read 2 more answers

A sample of a substance has a mass of 4.2 grams and a volume of 6 milliliters. The density of this substance is

Tanzania [10]

Hello!

Mass =4.2 g

Volume =6 mL

Therefore:

Density = mass / volume

Density = 4.2 / 6

Density = 0.7 g/mL

Hope that helps!

4 0

3 years ago

Read 2 more answers

Consider the following reaction and situations 1 through 10. In the spaces provided, clearly indicate the best response to each

olchik [2.2K]

Answer:

1. C. no change

2. A. increase

3. E. shift to the right

4. A. increase

5. E. shift to the right

6. A. increase

7. F. cannot be determined

8. B increase

9. D. shift to the left

10 F. cannot be determined

Explanation:

According to Le Chaterlier principle, when a reaction is in equilibrium and one of the constraints that affect reactions is applied, the equilibrium will shift so as annul the effects of the constraints.

From the equation: C(s) + H2O(g) ⇌ CO(g) + H2(g),

H is greater than 0, meaning that the system is endothermic, that is energy is absorbed.

1. If the pressure of the system is increased, there would be no change to the system because there are equal number of moles of products and reactants.

2. If H2 concentration is decreased, the equilibrium will shift to the right and more products will be formed. Hence, the concentration of CO will increase.

3. If H2 concentration is decreased, the equilibrium will shift to the right to annul the effects of the decrease in the concentration of a product.

4. If the concentration of H2 is increased, the equilibrium will shift to the left to annul the effects of increased concentration of a product. Hence, more H2O would be formed.

5. If H2 (a product) is removed, and C (a reactant) is added, more of the products will be formed in order to annul the effects of the actions. Hence, equilibrium will shift to the right.

6. If the amount of C (a reactant) is increased, the equilibrium will shift to the right. Hence, more H2 will be formed.

7. The reaction is endothermic, hence an increase in temperature will ordinarily shift the equilibrium to the right. However, the addition of H2 (a product) is supposed to shift the equilibrium to the left. Hence, the effects of simultaneous addition of the two actions become indeterminate.

8. Since the reaction is endothermic, increase in the temperature of the system will shift the equilibrium to the right. Hence, more CO will be formed.

9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

5 0

3 years ago

What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O

nadya68 [22]

Answer:

$V=43.46mL$

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

$3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O$

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

$n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4$

Then, given the molarity, it is possible to obtain the milliliters as follows:

$V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL$

Best regards!

7 0

3 years ago