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Delvig [45]
3 years ago
12

Write a balanced half-reaction for the reduction of permanganate ion (Mno) to manganese ion Mn?) in acidic aqueous solution. Be

sure to add physical state symbols where appropriate. 0-0 Cb X 5 2
Chemistry
1 answer:
Luden [163]3 years ago
6 0

Explanation:

Permanganate is general name for the chemical compound which containes manganate (VII) ion. Permanganate(VII) ion is strong oxidizing agent as  manganese is in +7 oxidation state and can be easily reduced and oxidize others.

The balanced half reaction for reduction of the permanganate ion, MnO_4^- to manganese ion, Mn^{2+} is shown below:

MnO_4^-_{(aq)}+8H^+_{(aq)}+5e^-\rightarrow Mn^{2+}_{(aq)}+4H_2O_{(l)}

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ziro4ka [17]

Answer:

1.9000kgm/s

2.258,000kgm/s

Explanation:

1.Momentum is given as the product of mass by velocity of an object.

Momentum,p=mv

m=1,500kh, v=6m/s

p=1500*6\\p=9000kgm/s

2.Momentum,p=mv

m=7800kg, v=30m/s

p_1=7800*30\\p_1=234,000kgm/s\\

new mass=7800+800=8600

As mass is increased, so does the resultant velocity as mass is directly proportional to velocity.

p_2=8600*30\\p_2=258,000kgm/s

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3 years ago
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3 years ago
The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P
Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

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To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is \rm PbS. There will be 890 grams of \rm PbS in one kilogram of this rock.

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M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}.

How many moles of \rm PbS formula units in that 890 grams of \rm PbS?

\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol.

There's one mole of \rm Pb in each mole of \rm PbS. There are thus \rm 3.71980\; mol of \rm Pb in one kilogram of this rock.

What will be the mass of that \rm 3.71980\; mol of \rm Pb?

m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg.

In other words, the \rm PbS in 1 kilogram of this rock contains \rm 0.770743\; kg of lead \rm Pb.

How many kilograms of the rock will contain enough \rm PbS to provide 1.5 kilogram of \rm Pb?

\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg.

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3 years ago
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Explanation:

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