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amid [387]
3 years ago
11

A solution of ammonium carbonate is mixed with a solution of magnesium nitrate. If magnesium carbonate is insoluble in water, wh

at is the reaction?​

Chemistry
1 answer:
Doss [256]3 years ago
3 0

Double displacement/ Double decomposition/ Metathesis

Explanation:

The insoluble magnesium carbonate produced is called a precipitate. In a double displacement reaction, ionic compounds combines to produce precipitates.

In this type of reaction, species are usually exchange to form new compounds.

One of the following is a driving force for double displacement reaction:

  • Formation of insoluble compounds called precipitates.
  • Formation of water or any other non - ionizing compound
  • liberation of a gaseous product.

learn more:

Precipitate brainly.com/question/8896163

#learnwithBrainly

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A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol
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Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

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q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the solution

c_1 = specific heat of calorimeter = 12.1J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water or solution = Density\times Volume=1/mL\times 100.0mL=100.0g

\Delta T = change in temperature = T_2-T_1=(26.3-20.2)^oC=6.1^oC

Now put all the given values in the above formula, we get:

q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]

q=2623.61J

Now we have to calculate the enthalpy change for the process.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = Concentration\times Volume=1M\times 0.050L=0.050mole

\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

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Which of the following correctly describes the difference (on average) between an electron in a 2s orbital and an electron in a
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