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2 years ago

List the sequence of events that occurred in the area represented by the profile above.

Chemistry

1 answer:

Fittoniya [83]2 years ago

7 0

1. Igneous intrusion Y formed.

2. The shale layer was deposited.

3. Igneous intrusion X formed.

4. The limestone layer was deposited.

5. The mudstone layer was deposited.

B 1. The mudstone layer was deposited.

2. The limestone layer was deposited.

3. The shale layer was deposited.

4. Igneous intrusion Y was forced onto the mudstone, limestone, and shale.

5. Igneous intrusion X was forced onto the mudstone, limestone, and shale.

C 1. Igneous intrusion X formed.

2. Igneous intrusion Y formed.

3. The shale layer was deposited.

4. The limestone layer was deposited.

5. The mudstone layer was deposited.

D 1. The layer of mudstone was deposited.

2. The layer of limestone was deposited.

3. Igneous intrusion Y was forced onto the mudstone and limestone.

4. The shale layer was deposited.

5. Igneous intrusion X was forced over the mudstone, limestone, and shale

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3 years ago

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

Explanation:

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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Explanation:

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