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Rasek [7]
3 years ago
15

To begin the experiment, a bomb calorimeter is filled with 1.11 g CH4 and an excess of oxygen. The heat capacity of the calorime

ter, including the bomb and the water, is 4.319 kJ g C⋅°. The initial temperature of the system was 24.85°C, and the final temperature was 35.65°C.
Using the formula ∆H =−M ⋅ C ⋅∆T , solve for the heat of combustion of. 1.11 g CH4.
Chemistry
1 answer:
Flura [38]3 years ago
6 0
There are several information's already given in the question. Based on those information's the answer can be easily determined.

M = <span>1.11 g CH4
C = </span>4.319 kJ g C⋅°
∆T = 35.65 - 24.85 degree centigrade
     = 10.8 degree centigrade.
Then
∆H =−M ⋅ C ⋅∆T
      = - 1.11 * 4.319 * 10.8
      = - 51.776 kJ/<span>mol

I hope the procedure is clear enough for you to understand.</span>
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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

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

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

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

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 = 0°C

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

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

8 0
3 years ago
The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from
balu736 [363]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final volume of the gas is 100 ml.</em></u>

Explanation:

 Given:

Initial pressure (P_{1}) = 600 mm of Hg

Final pressure (P_{2}) = 1200 mm of Hg

Initial volume (V_{1}) = 200 ml      

To find:

Final volume (V_{2})

We know;

According to the ideal gas equation,

    P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

 From the above mentioned equation,

        P × V = constant

\frac{P_{1} }{P_{2} } = \frac{V_{1} }{V_{2} }

Where,

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1})  represents the initial volume of the gas

(V_{2})  represents the final volume of the gas

So;

\frac{600}{1200} = \frac{V_{2} }{200}    

V_{2} = 100 ml

<u><em>Therefore the final volume of the gas is 100 ml.</em></u>                                                                                                                                                                              

5 0
3 years ago
Why is monatomic compound nonsense?
Tamiku [17]
<span>Because mona means 1 atom, and a compound is made up of 2 or more atoms.</span>
3 0
3 years ago
Please help me with this question
natka813 [3]

Answer:

B

Explanation:

3 0
2 years ago
What volume of a 0.550 M solution of potassium hydroxide (KOH) can be made with 19.9 g of potassium hydroxide?
Natalija [7]

Answer:

0.645 L

Explanation:

To find the volume, you need to (1) convert grams to moles (using the molar mass) and then (2) calculate the volume (using the molarity ratio). The final answer should have 3 sig figs to match the sig figs of the given values.

(Step 1)

Molar Mass (KOH): 39.098 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (KOH): 56.104 g/mol

19.9 grams KOH              1 mole
--------------------------  x  -----------------------  =  0.355 moles KOH
                                     56.014 grams

(Step 2)

Molarity = moles / volume                            <----- Molarity ratio

0.550 M = 0.355 moles / volume                 <----- Insert values

(0.550 M) x volume = 0.355 moles              <----- Multiply both sides by volume

volume = 0.645 L                                          <----- Divide both sides by 0.550

6 0
1 year ago
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