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3 years ago

6

2C4H10+ 13O2→ 8CO2+ 10H2O

1 answer:

katen-ka-za [31]3 years ago

8 0

Answer:

Combustion reaction

Explanation:

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The electron affinity of Bi is -94.6 kJ and the value for At is -280 kJ. What is the approximate electron affinity for Po?

lys-0071 [83]

Answer:

48 KJ mol-1

Explanation:

3 0

3 years ago

In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?

Oksanka [162]

Answer:- C. H

Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.

As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In $H_2O$ , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in $Cl^-$ is -1. On product side, the oxidation number of hydrogen in $H_2$ is zero and in $OH^-$ the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in $Cl_2$ is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

Oxidation number of H is changing from +1 to 0 which is reduction.

So, the right choice is C.H

8 0

3 years ago

Help pls :) I am stuck on this chemistry question about percentage yields!

Charra [1.4K]

You just switch them around

4 0

3 years ago

An atom requires 6.24 × 10−19J to remove its furthest electron in the gas phase. What is the maximum wavelength of electromagnet

daser333 [38]

<span>Report your numerical answer in units of nm. Use significant figur</span>

3 0

3 years ago

Read 2 more answers

Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,

7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 ) $N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$ ; Kc = 1.54e - 31

2) $N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)$ ; Kc = 2.16e - 24

upon reversing ( 2 ) equation

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$ Kc = 1/2.16e - 24

now adding 1 and reversed equation (2)

$N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)$

$N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)$

we get ,

$N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)$ Kc = 1.54e-31 × 1/2.61e - 24

equilibrium constant of equation (3) is -

Kc = 1.54e - 31 / 2.61e - 24

3 0

3 years ago