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Wittaler [7]
2 years ago
15

Select the correct balanced oxidation-reduction reaction. Group of answer choices 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+

(aq) + 2Cr +(aq) + 7H2O(l) 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) + 2e- 7H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l) 14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
Chemistry
1 answer:
bezimeni [28]2 years ago
6 0

Answer:

The correct balanced oxidation- reduction reaction is:

14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

Explanation:

In this reaction, iron (Fe2+) is the reducing agent while Cr2O7^2- is the oxidizing agent.

The ion transfer is represented as shown below:

6 Fe 2+  - 6e- -----------> 6 Fe 3+          (oxidation)

2 Cr^6  + 6e^-  ----------> 2 Cr^3           (reduction)

From the unbalanced reaction

Fe2+ + Cr2O72- + H+ ---------->  Fe3+ + Cr3+ + H2O we will follow these steps to balance the reaction.

Step 1: break the equation into two half reactions stating which is oxidized and reduced.

Step 2: Balance the atoms on each sides; the hydrogen, oxygen

Step 3: Balance the gain also

Step 4: Give the electron gained on one side to be equal to the electron lost on the other side.

Step 5: Add the two half reactions and simplify the equation.

Doing this, we obtain

14H+(aq) + 6Fe2+(aq) + Cr2O72-(aq) -> 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

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Name of th molecule <br> 1. CH3CH2CHClCHBrCH3<br> 2.C=C-CH3<br> CH3CH=CHCH2
Grace [21]

Answer:

1: 2-bromo-3-chloropentane

Explanation:

find longest carbon chain =5

place the Br and Cl on the carbon chain

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4 0
2 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}
3 0
2 years ago
Gas A effuses 0.68 times as fast as Gas B. If the molar mass of Gas B is 17 g, what is the mass of Gas A?
BartSMP [9]
The mass of gas A is 11.56. i got this answer by 0.68 multiplied by 17.  because in this problem the key word is times.
4 0
3 years ago
Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at
kozerog [31]

Answer:

20 g/mol

Explanation:

We can use <em>Graham’s Law of diffusion</em>:

The rate of diffusion (<em>r</em>) of a gas is inversely proportional to the square root of its molar mass (<em>M</em>).

r = \frac{1 }{\sqrt{M}}

If you have two gases, the ratio of their rates of diffusion is

\frac{r_{2}}{r_{1}} = \sqrt{\frac{M_{1}}{M_{2}}}

Squaring both sides, we get

(\frac{r_{2}}{r_{1}})^{2} = \frac{M_{1}}{M_{2}}

Solve for <em>M</em>₂:

M_{2} = M_{1} \times (\frac{r_{1}}{r_{2}})^{2}

M_{2} = \text{39.95 g/mol} \times (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^{2}= \text{39.95 g/mol} \times (0.711 )^{2}

= \text{39.95 g/mol} \times 0.506 = \textbf{20 g/mol}

7 0
3 years ago
Say I grab a bunch of aluminum foil (9.5g) and ball it up. Aluminum has a specific heat of .9J/g*C. How much would the aluminum’
Brrunno [24]

Answer:

ΔT = 0.78 °C

Explanation:

Given data:

Mass of Al = 9.5 g

Specific heat capacity of Al = 0.9 J/g.°C

Temperature change = ?

Heat added = 67 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

67 J = 9.5 g × 0.9 j/g.°C × ΔT

67 J = 85.5 j/°C × ΔT

ΔT = 67 J  /  85.5 j/°C

ΔT = 0.78 °C

7 0
3 years ago
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