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Rzqust [24]
3 years ago
13

SnO2 + 2H2 → Sn + 2H2O

Chemistry
1 answer:
anzhelika [568]3 years ago
7 0

Answer:- B: 4.21 is the correct option.

Solution:- The given balanced equation is:

SnO_2+2H_2\rightarrow Sn+2H_2O

From the balanced equation, there is 1:1 mol ratio between SnO_2 and Sn.

To calculate the moles of SnO_2 required to produce500.0 grams of Sn, we need to convert the grams to moles and multiply it by the mole ratio as:

500.0gSn(\frac{1moleSn}{118.71gSn})(\frac{1moleSnO_2}{1moleSn})

= 4.21moleSnO_2

So, the correct option is B: 4.21 moles.

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ASAP MULTIPLE CHOICE WILL MARK BRAINLIEST
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How are plate tectonics and the rock cycle connected
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Answer:

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8 0
2 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
sammy [17]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

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

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

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

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

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

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

5 0
3 years ago
I need help with these
nikklg [1K]
2. Rubidium
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8 0
2 years ago
A piston is pressurized to 15.5 psi at 405 K. If the piston compresses the air, from 8.98 L to 7.55 L, and the pressure drops to
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Answer:

239.45 K

Explanation:

Ideal gas law formula is P1V1T2=P2V2T1

Rearrange that to get...

T2=T1P2V2/P1V1

Fill in the values and solve.

7 0
3 years ago
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