What is the density of object B? Does it sink or float in water?

2Na2O2 + 2CO2 → 2Na2CO3 + O2

ahrayia [7]

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield x 100%

81.0 % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

7 0

3 years ago

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A student dissolves 15.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.13/gmL. The student notices that the vo

Kruka [31]

Answer: The molarity and molality of sucrose solution is 0.146 M and 0.129 m respectively

Explanation:

Calculating the molarity of solution:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sucrose = 15 g

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 300 L

Putting values in above equation, we get:

$\text{Molarity of sucrose solution}=\frac{15\times 1000}{342.3\times 300}\\\\\text{Molarity of sucrose solution}=0.146M$

Hence, the molarity of sucrose solution is 0.146 M

Calculating the molality of solution:

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

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

Density of solvent = 1.13 g/mL

Volume of solvent = 300 mL

Putting values in above equation, we get:

$1.13g/mL=\frac{\text{Mass of solvent}}{300mL}\\\\\text{Mass of solvent}=(1.13g/mL\times 300mL)=339g$

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

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (sucrose) = 15 g

$M_{solute}$ = Molar mass of solute (sucrose) = 342.3 g/mol

$W_{solvent}$ = Mass of solvent = 339 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{15\times 1000}{342.3\times 339}\\\\\text{Molality of solution}=0.129m$

Hence, the molality of sucrose solution is 0.129 m

5 0

4 years ago

Your lab partner combined chloroform and acetone to create a solution where the mole fraction of chloroform, Xchloroform, is 0.1

jeyben [28]

Answer:

Explanation:

[u]Assumptions[/u]

1. There is exactly 1 mole of chloroform

2. The liquids mix together well such that the volume of the solution is the sum of the volumes of the two liquids

Given that the mole fraction of the Chloroform is 0.171

Mole fraction of Chloroform =

Mole of chloroform /(Mole of chloroform + mole of acetone)

According to assumptions, mole of chloroform is equal to 1

Therefore 0.171 =1/(1+mole of acetone)

1 + mole of acetone = 1/0.171

Mole of acetone = 1(/0.171) - 1

Moles of acetone = 4.85mol.

From Stochiometry

Mass of acetone = Mole of acetone * Molar Mass of acetone

Molar mass of acetone = 58.1grams/mol

Mass of acetone = 4.85 *58.1 = 282g =0.282kg

Mass of chloroform = moles of chloroform *Molar mass of Chloroform

Molar mass of chloform = 119.4 grams/mol

Mass of chloroform = 1* 119.4 =119.4g=0.1994kg

Volume of acetone = Mass of acetone / Density of acetone

Volume of acetone = 282/0.791

Volume of acetone = 357mL

Volume of Chloroform = Mass of Chloroform /Density of Chloroform

Volume of Chloroform = 119.4/1.48

= 81mL

Total volume of solution = 357mL+81mL = 438mL = 0.438L

1. Molarity = Moles of solute(chloroform)/mass of solvent (acetone) in kg

Molarity = 1/0.282 = 3.55molal

2. Molarity = moles of solute( chloroform) /Volume of solution

= 1/0.438 =2.28Molar

Therefore the molality and molarity respectively are 3.55 and 2.28.

4 0

4 years ago

How does the energy provided by the Sun affect global ocean currents?

777dan777 [17]

The answer is C. The sun heats waters near the equator, which then move toward the polls where the waters cool.

8 0

3 years ago

What is the chemical formula for aluminum fluoride?

algol13

Answer:

AlF3

Explanation:

Aluminum fluoride is an inorganic compound of chemical formula AlF3. Such a compound originates from the treatment of aluminum hydroxide particles with hydrogen fluoride, which react and give rise to aluminum fluoride. This is a very important solid and crystalline element during the production of other compounds that use aluminum oxides because it has the ability to lower the melting point.

3 0

4 years ago

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