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Fittoniya [83]
3 years ago
10

How many joules of energy do you have if you have 0.45 L of water in a can and you heat the 23C water to 55C site:socratic.org

Chemistry
1 answer:
Bess [88]3 years ago
4 0

Answer:

Q = 60192 j

Explanation:

Given data:

Volume of water = 0.45 L

Initial temperature = 23°C

Final temperature = 55°C

Amount of heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  55°C - 23°C

ΔT = 32°C

one L = 1000 g

0.45 × 1000 = 450 g

Specific heat capacity of water is 4.18 j/g°C

Q = m.c. ΔT

Q = 450 g.  4.18 j/g°C.  32°C

Q = 60192 j

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fgiga [73]

<u>Answer:</u> The pH of the solution is 12.61

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}       ......(1)

  • <u>For hydrazoic acid:</u>

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol

  • <u>For NaOH:</u>

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol

The chemical reaction for hydrazoic acid and NaOH follows the equation:

                   HN_3+NaOH\rightarrow NaN_3+H_2O

<u>Initial:</u>           0.291        0.342

<u>Final:</u>                  0          0.051                 0.291      0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L    (Conversion factor:  1 L = 1000 mL)

  • <u>For NaOH left:</u>

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

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

pOH=-\log[OH^-]

We are given:

[OH^-]=0.0408M

Putting values in above equation, we get:

pOH=-\log(0.0408)\\\\pOH=1.39

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

pH+pOH=14\\\\pH=14-1.39=12.61

Hence, the pH of the solution is 12.61

8 0
3 years ago
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