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zzz [600]
3 years ago
12

The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

he enthalpy change for this reaction using the following enthalpy changes of formation. AH! [Cu(NO3)2) = -302.9 kJ mol? AH, (CuO) = -157.3 kJ mol?. AH[NO2) = +33.2 kJ mol.
Chemistry
1 answer:
Basile [38]3 years ago
6 0

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?

We have :

Enthalpy changes of formation of following s:

\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol

\Delta H_{f,CuO}=-157.3 kJ/mol

\Delta H_{f,NO_2}= 33.2 kJ/mol

\Delta H_{f,O_2}= 0 kJ/mol (standard state)

\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]

The equation for the enthalpy change of the given reaction is:

\Delta H_{rxn} =

=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})

\Delta H_{rxn}=

(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)

\Delta H_{rxn}=424 kJ

The enthalpy change for the given reaction is 424 kJ.

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babymother [125]

Answer:

16.56 g

Explanation:

Mass is the production of Volume and Density.

m = V. d = 6 × 2.76 = 16.56 g

6 0
3 years ago
You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0
3 years ago
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The equation 2al(s) + 3br2(l) → 2albr3(s) is a(n) ______________ reaction.
Svetach [21]
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How many liters of nitrogen dioxide are in 16 g of NO2 at STP?
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Answer: 23 liter

Explanation: Im pretty sure

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Find the kinetic energy of a boy of mass 6.5kg running at a velocity of 6.0m/s​
ololo11 [35]

Explanation:

Given,

M=6.5kg,velocity(v)=6m/s

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=1/2×6.5×6^2

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=18×6.5

=117 J

3 0
2 years ago
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