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Andrej [43]
3 years ago
5

What substance is reduced in the reaction 2Fe 2+ + CI2 = 2Fe 3+ + 2CI- ?

Chemistry
1 answer:
atroni [7]3 years ago
5 0

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

Hence, $Fe^{2+}$ is oxidized and $Cl_{2}$ is reduced

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(Atomic masses used are C = 12.0 g mol–1, O = 16.0 g mol–1, H = 1.01 g mol–1, Cu = 63.5 g mol–1,)
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<u>Question 1</u>

The formula mass of methanol is 12.0+3(1.01)+16.0+1.01=32.04 g/mol.

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<u>Question 2</u>

The formula mass of copper(II) carbonate is 63.5+12.0+3(16.0)=123.5 g/mol.

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<u>Question 3</u>

The formula mass of cyclohexane is 12.0(6)+1.01(12)=84.12 g/mol.

So, 3.0 grams of cyclohexane is the same as 3.0/84.12=\boxed{0.036} moles.

<u>Question 4</u>

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<u>Question 5</u>

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A 2.00-mol sample of hydrogen gas is heated at constant pressure from 294 K to 414 K. (a) Calculate the energy transferred to th
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Answer:

a) The energy transferred is 6.91 kJ

b) The internal energy is 4.90 kJ

c) The work done on the gas is - 2.01 kJ

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 2.00 moles

Pressure = constant

Temperature is heated from 294 K to 414 K

Molar heat capacity of hydrogen gas = 28.8 J/mol*K

Step 2: Calculate the energy transferred to the gas by heat.

Q = n* Cp * ΔT

⇒with Q =the energy transferred

⇒with n = the number of moles = 2.00 moles

⇒with Cp = the Molar heat capacity of hydrogen gas = 28.8 J/mol*K

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Q = 2.00 * 28.8 * 120

Q = 6912 J = 6.91 kJ

Step 3: Calculate the increase in its internal energy.

ΔEint = n*Cv*ΔT

⇒with ΔEint = the increase in its internal energy.

⇒with n = the number of moles = 2.00 moles

⇒with Cv = The constant volume = 20.4 J/mol*K

⇒with  ΔT = Temperature 2 - Temperature 1 = 414 - 294 = 120K

ΔEint = 2.00 * 20.4 * 120

ΔEint =4896 J = 4.90 kJ

Step 4: Calculate the work done on the gas.

Work done on the gas = -Q + ΔEint

W = -6.91 kJ + 4.90 kJ

W = -2.01 kJ

6 0
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