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oee [108]
3 years ago
14

State with reasons, whether sulphur dioxide is acting as an oxidizing agent or a reducing agent in each of the following reactio

ns:
•2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s)

•SO2(g) +H2O(l) +NaClO(aq) -> NaCl(aq) + H2SO4(aq)
Chemistry
1 answer:
evablogger [386]3 years ago
4 0

Answer:

A) oxidizing agent is SO2

B) NaClO is the oxidizing agent

Explanation:

A) This is a redox reaction in which oxidation and reduction occur simultaneously.

Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);

H2S is reduced as follows;

H2S → S + 2H+ + 2e−

We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.

B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)

In this, SO2 undergoes oxidation and NaClO is the oxidizing agent

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Two atoms of calcium and one molecule of oxygen gas react and yield two molecules of calcium oxide.
Vitek1552 [10]
2Ca + O2 ——————-> 2CaO2
3 0
3 years ago
Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802
Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

3 0
2 years ago
What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?
zmey [24]

Answer:

228 mL

Explanation:

M1*V1 = M2*V2

M1 = 6.58 M

V1 = ?

M2 = 3.00 M

V2 = 500 mL

V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL

3 0
3 years ago
1. Analysis of an unknown substance formerly used in rocket fuel reveals a composition of 93.28% nitrogen and 6.72% hydrogen by
Nitella [24]

Answer:

The formula of the compound is:

N2H2

Explanation:

Data obtained from the question:

Nitrogen (N) = 93.28%

Hydrogen (H) = 6.72%

Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:

N = 93.28%

H = 6.72%

Divide by their molar mass

N = 93.28 /14 = 6.663

H = 6.72 /1 = 6.7

Divide by the smallest

N = 6.663 / 6.663 = 1

H = 6.72 /6.663 = 1

Therefore, the empirical formula is NH.

Now, we can obtain the formula of the compound as follow:

The formula of a compound is simply a multiple of the empirical formula.

[NH]n = 30.04

[14 + 1]n = 30.04

15n = 30.04

Divide both side by 15

n = 30.04/15

n = 2

Therefore, the formula of the compound is:

[NH]n => [NH]2 => N2H2

6 0
3 years ago
2AgNO3 (aq) + BaCl2 (aq)<br> --&gt;<br> 2AgCl (s) + Ba(NO3)2 (aq)
dimaraw [331]

2AgCl (s) + Ba(NO3)2 (aq)

6 0
3 years ago
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