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kow [346]
3 years ago
10

Calculate the enthalpy change, ΔH, for the process in which 31.6 g of water is converted from liquid at 3.2 ∘C to vapor at 25.0°

C . For water, ΔHvap = 44.0 kJ/mol at 25.0°C and Cs = 4.18 J/(g*°C) for H2O(l).
Chemistry
1 answer:
amm18123 years ago
5 0

Answer:

Enthalpy change = 74.36 kJ

Explanation:

Enthalpy change is defines as the heat absorbed or evolved during a chemical reaction at constant pressure and volume

Reaction:

H2O(l) --> H2O(g)

DHr = mCsDT

Where m is the mass of water

Cs is the specific heat capacity

DT is the temperature difference

= 31.6 * 4.18 * (25 - 3.2)

= 2879.518 J

DHvap = 44kJ/mol

Moles of water = mass/molecular weight

Molecular weight = (1*2) + 16

= 18g/mol

Moles of water = 31.6/18

= 1.756 moles

DHvap = 44 * 1.756

= 77.244kJ

DH = DHproduct - DHreactant

DHproduct = DHvap = 77.24kJ

DHreactant = DHr = 2879.518J = 2.880kJ

= 77.24 - 2.880

DH = 74.36kJ

Enthalpy change = 74.36 kJ

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8 0
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A solution was found to have a 15.6 % transmittance at 500 nm, its wavelength of maximum absorption, using a cell with a path le
Gnesinka [82]

For the absorbance of the solution in a 1.00 cm cell at 500 nm  is mathematically given as

A’ = 0.16138

<h3>What is the absorbance of the solution in a 1.00 cm cell at 500 nm?</h3>

Absorbance (A) 2 – log (%T) = 2 – log (15.6) = 0.8069

Generally, the equation for the Beer’s law is mathematically given as

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0.8069 = ε*c*(5.00 )

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then for when ε*c is constant

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7 0
2 years ago
You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 2
Katyanochek1 [597]

<u>Answer:</u> The mass of acetic acid used is 0.12 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is CH_3COOH

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL

Putting values in above equation, we get:

1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\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 acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g

Hence, the mass of acetic acid used is 0.12 grams

5 0
3 years ago
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