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wolverine [178]
3 years ago
13

When 0.103 g of Zn (s) is combined with enough HCl to make 50.0 mL of solution in a coffee cup calorimeter, all of the zinc reac

ts, raising the temperature of the solution from 22.5 degree C to 23.7 degree C. Find Delta H_rxn per mole of Zn. (Use 1.02 glmL for the density of the solution and 4.18 J/g degree C for the specific heat capacity.)
Zn (s) + 2 HCl (aq) rightarrow ZnCl_2 (aq) + H_2 (g)
Delta H_rxn/mol Zn = ____kJ/mol.
Chemistry
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

\Delta H_{rxn}=162940 J/mol

Explanation:

To calculate the heat of reaction we need to determine:

  1. The heat produced
  2. The moles of Zn that reacted

<u>Moles of zinc (Zn)</u>

It says that the 0.103 g reacted completely, knowing that the Zn molecular weight is 65.4 g/mol:

n_{Zn}=\frac{0.103g}{65.4 g/mol}=1.57*10^{-3}mol

<u>Heat released by the reaction</u>

The solution goes from 22.5 °C to 23.7°C:

Q=Cp*\rho *V*(T_f - T_i)

Q=4.18 \frac{J}{g* ^{\circ}C}*1.02 g/mL*50mL*(23.7-22.5)^{\circ}C

Q=255.8 J

<u>Heat of reaction</u>

\Delta H_{rxn}=\frac{-Q}{n_{Zn}}

\Delta H_{rxn}=\frac{-255.8 J}{1.57*10^-3}}=162940 J/mol

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For the following reaction,
jeka94

Answer:

The answer to your question is:

1.- CO

2.- 0.414 moles of CO2

Explanation:

Data

                               2CO   + O2      ⇒    2CO2

CO = 0.414 moles

O2 = 0.418

 

Process

theoretical ratio   CO/O2 = 2/1 = 1

experimental ratio  CO/O2 = 0.414/0.418 = 0.99

Then the limiting reactant is CO

2.-

                    2 moles of CO ---------------  2 moles of CO2

                    0.414 moles of CO ---------  x

                   x = (0.414 x 2) / 2

                   x = 0.414 moles of CO2

                   

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Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

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[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

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This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

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A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

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