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3 years ago

How many grams of Sn will be created when 114 grams of SnO2 react with excess H2? SnO2 + H2 ->Sn + H2O

Chemistry

1 answer:

Snezhnost [94]3 years ago

5 0

Explanation:

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How many major layers of atmosphere does earth have

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Five major.

From lowest to highest, the major layers are the troposphere, stratosphere, mesosphere, thermosphere and exosphere. Troposphere.

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2 years ago

A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL

AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

Answer: The molecular formula for the given organic compound is $C_6H_{14}O_2$

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=i\times K_f\times m$

Or,

$\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal boiling point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute = 10.20 g

$M_{solute}$ = Molar mass of solute = ?

$W_{solvent}$ = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

$(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g$

Calculating the molecular formula:

We are given:

Percentage of C = 60.98 %

Percentage of H = 11.94 %

Percentage of O = (100 - 60.98 - 11.94) % = 27.08 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 60.98 g

Mass of H = 11.94 g

Mass of O = 27.08 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{60.98g}{12g/mole}=5.082moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.94g}{1g/mole}=11.94moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.08g}{16g/mole}=1.69moles$

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 1.69 moles.

For Carbon = $\frac{5.082}{1.69}=3$

For Hydrogen = $\frac{11.94}{1.69}=7.06\approx 7$

For Oxygen = $\frac{1.69}{1.69}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 7 : 1

The empirical formula for the given compound is $C_3H_7O$

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 = 118.2 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

$n=\frac{118.2g/mol}{59g/mol}=2$

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

$C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2$

Hence, the molecular formula for the given organic compound is $C_6H_{14}O_2$

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3 years ago

A stock solution is made by dissolving 66.05 g of (NH4)2SO4 in enough water to make 250 mL of solution. A 10.0 mL sample of this

Degger [83]

If the solute is properly distributed in the given volume, there are 2.642 g of (NH4)2SO4 per 10 mL. For the new solution, divide the 2.642 g by the molar mass of the compound. The answer is 0.02 moles. Then, divide this by the new volume, 50 mL or 0.05 L. The concentration of the new solution is 0.4 M.

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3 years ago

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What is the difference between ionic formula and the molecular formula?

Kaylis [27]

Answer:

Molecular compounds are pure substances formed when atoms are linked together by sharing of electrons while ionic compounds are formed due to the transfer of electrons. 2. ... Molecular compounds are formed between two non-metals while ionic compounds are formed between metals and non-metals.

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3 years ago

1. What is the equation for Boyle's Law?.

Sveta_85 [38]

Answer: 1. P1V1 = P2V2

2. P stands for pressure

3. Units for Pressure are atm and Pa

4. V stands for volume

5. Units for volume is in mL

Explanation: Boyle's Law is a gas law that states the relationship between pressure and volume of a gas.

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3 years ago

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