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Jlenok [28]
2 years ago
11

5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) ⟶ 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) determine if each of the following statements is True or Fa

lse. 1) MnO4−(aq) is the oxidizing agent. 2) Fe2+(aq) gains electrons. 3) Mn2+(aq) is the reducing agent. 4) H+(aq) undergoes oxidization.
Chemistry
1 answer:
aalyn [17]2 years ago
3 0

Explanation:

Fe2+ lose electrons to become Fe 3+ hence oxidized

maO4- is the oxidizing agent

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The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
Katen [24]

Answer:

The pH of the solution is 8.

Explanation:

To which options are correct, let us determine the concentration of the hydroxide ion, [OH-] and the pH of the solution. This is illustrated below:

1. The concentration of the hydroxide ion, [OH-] can be obtained as follow:

pOH = –Log [OH-]

pOH = 6

6 = –Log [OH-]

–6 = Log [OH-]

[OH-] = Antilog (–6)

[OH-] = 1x10^–6 mol/L

2. The pH of the solution can be obtained as follow:

pH + pOH = 14

pOH = 6

pH + 6 = 14

pH = 14 – 6

pH = 8.

From the calculations made above,

[OH-] = 1x10^–6 mol/L

pH = 8.

Therefore, the correct answer is:

The pH of the solution is 8

3 0
2 years ago
The plastic straws were placed under the wooden block to
Sveta_85 [38]
Answer: C
I hope this helped you
6 0
2 years ago
Strontium nitrate, Sr(NO3)2, is used in fireworks to produce brilliant red colors, Suppose we need to prepare 366.6 ml. of 0.115
natulia [17]

Answer:

\boxed{\text{8.91 g}}

Explanation:

1. Calculate the moles of Sr(NO₃)₂

n = \text{366 mL} \times \dfrac{\text{0.115 mmol}}{\text{1 mL}}= \text{42.09 mmol}

2. Calculate the mass of SrNO₃)₂

m = \text{42.09 mmol} \times \dfrac{\text{211.63 mg}}{\text{1 mol}}= \text{8910 mg} = \text{8.91 g}\\\text{The mass of strontium nitrate required is }\boxed{\textbf{8.91 g}}

3 0
2 years ago
The rate of effusion of nitrogen gas (N2) is 1.253 times faster than that of an unknown gas. What is the molecular weight of the
ollegr [7]

Answer:

43.96

Explanation:

Graham's law was applied and the rates of effusion of nitrogen and the unknown gas were compared as shown in the image. The unknown gas is heavier than hydrigen hence it effuses slower than hydrogen as anticipated by Graham's law.

5 0
2 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
2 years ago
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