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m_a_m_a [10]
1 year ago
14

A 2.83 g piece of zinc (density = 7.14 g/ml) is added to a graduated cylinder that contains 12.13 ml H2O. What will be the final

volume reading on the graduated cylinder, in ml?
Chemistry
1 answer:
Maslowich1 year ago
3 0

The final volume reading on the graduated cylinder is 12.52ml

Density = mass / volume

Volume = mass / density

Volume of zinc = 2.83/7.14

Volume of zinc = 0.3963

Initial volume = 12.13 ml

Total volume = initial volume+ volume of zinc

Total volume = 12.13 + 0.3963

Total volume = 12.5263 = 12.52 ml

To know more about volume to go the given link :

brainly.com/question/11510153

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<u>Answer:</u> The enthalpy change of the reaction is -27. kJ/mol

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

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

Density of water = 1 g/mL

Volume of water = 25.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{25.0mL}\\\\\text{Mass of water}=(1g/mL\times 25.0mL)=25g

To calculate the heat released by the reaction, we use the equation:

q=mc\Delta T

where,

q = heat released

m = Total mass = [1.25 + 25] = 26.25 g

c = heat capacity of water = 4.18 J/g°C

\Delta T = change in temperature = T_2-T_1=(21.9-25.8)^oC=-3.9^oC

Putting values in above equation, we get:

q=26.25g\tiimes 4.18J/g^oC\times (-3.9^oC)=-427.9J=-0.428kJ

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

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

Given mass of ammonium nitrate = 1.25 g

Molar mass of ammonium nitrate = 80 g/mol

Putting values in above equation, we get:

\text{Moles of ammonium nitrate}=\frac{1.25g}{80g/mol}=0.0156mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = -0.428 kJ

n = number of moles = 0.0156 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{-0.428kJ}{0.0156mol}=-27.44kJ/mol

Hence, the enthalpy change of the reaction is -27. kJ/mol

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3 years ago
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