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oksano4ka [1.4K]
3 years ago
13

A beaker contains 0.50 mol of potassium bromide in 600 ml of water. an additional 600 ml of water is added. the number of moles

of potassium bromide in the beaker is
Chemistry
1 answer:
abruzzese [7]3 years ago
5 0
The answer is 0.50 mol. The number of moles of potassium bromide in the beaker will not change since there is no additional potassium bromide introduced to the solution. The molarity of potassium bromide, the number of moles of water, the volume of the water solvent and the volume of the solution will all change if water is added.
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Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa
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Answer:

T final = 80°C

Explanation:

  • Q = mCpΔT

∴ Q = 18000 cal

∴ m H2O = 300 g

∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K

∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

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Consider the following intermediate chemical equations.
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Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)  \Delta H_2=3438kJ (2)

Net chemical equation:

PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H=? (3)

Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2 \Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ   (4)

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