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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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A substance in a specific state of matter was transferred from a cylindrical shaped container to a cube shaped container. The su
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D. The particles move up and down without changing their position

Explanation:

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In aqueous solution, hypobromite ion, BrO-, reacts to produce bromate ion, BrO3 -, and bromide ion, Br-, according to the follow
maw [93]

Answer:

22.73s

Explanation:

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the initial concentration of BrO- [A]o = 0.80 M

time = ?

Final concentration [A]t= one-half of 0.80 M = 0.40M

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5 0
3 years ago
The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use EscE 0.241V.
tino4ka555 [31]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  value is   K_b  =   1.89 *10^{-6}

Explanation:

From the question we are told that

   The  voltage of the cell is  V  =  0.731 \  V

Generally K_b is mathematically represented as  

           K_b  =  \frac{K_w }{ K_a }

Where  K_w  is the equilibrium constant for this auto-ionization of water with a value  K_w  =  1.0 *10^{-14}

Generally the E_{cell} is mathematically represented as

       E_{cell} =  V  -  E_{SCE}

=>     E_{cell} =  0.731 - 0.241

=>       E_{cell} =   0.49 V

This  E_{cell} is mathematically represented as

             E_{cell} =  \frac{0.0592}{n} *  log K_a

Where n is the number of moles which in this question is  n = 1

         So  

         0.490 =  \frac{0.0592}{1}  *  log K_a

=>      K_a  =  5.30*10^{-9}

So  

     K_b  =  \frac{ K_w}{ K_a}

=>   K_b  =  \frac{1.0 *10^{-14}}{ 5.30*10^{-9}}

=>    K_b  =   1.89 *10^{-6}

5 0
3 years ago
A student needs to prepare a stock solution for the lab - 500.0 ml of 0.750 m nitric acid. The student is provided with concentr
dlinn [17]

Molarity of concentrated nitric acid = 15.9 M

Volume of the stock solution to be prepared = 500.0 mL

Concentration of the stock that is to be prepared = 0.750 M

Calculating moles from molarity and volume of stock:

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Calculating volume of concentrated nitric acid to be taken for the preparation of stock solution:

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Converting L to mL:

0.0236L*\frac{1000mL}{1L} = 23.6 mL

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Therefore, 976.4 mL distilled water is to be added to 23.6 mL of 15.9 M nitric acid solution to prepare 500.0 mL of 0.750M nitric acid.

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