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olchik [2.2K]
3 years ago
12

Plz answer fasttt

Chemistry
1 answer:
wlad13 [49]3 years ago
5 0

Answer:

I believe the answer would be C4H9O2.

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Assume reaction N (g )space plus O (g )rightwards arrow N O (g )occurs in high layers of the atmosphere. What is the H subscript
8090 [49]

Answer:

Kindly check the explanation section.

Explanation:

From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.

In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;

∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).

That is the heat of formation of products minus the heat of formation of the reaction g specie(s).

Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.

−382.185 = x - 472.435kj/mol = 90.25 kJ/mol

5 0
3 years ago
If 2.0×10−4 moles of S2O2−8 in 170 mL of solution is consumed in 170 seconds , what is the rate of consumption of S2O2−8?
Ostrovityanka [42]

Answer : The rate of consumption of S_2O_2^{-8} is, 7.0\times 10^{-6}M/s

Explanation : Given,

Moles of S_2O_2^{-8} = 2.0\times 10^{-4}mol

Volume of solution = 170 mL = 0.170 L     (1 L = 1000 mL)

Time = 170 s

First we have to calculate the concentration.

\text{Concentration}=\frac{\text{Moles}}{\text{Volume of solution}}

\text{Concentration}=\frac{2.0\times 10^{-4}mol}{0.170L}

\text{Concentration}=1.2\times 10^{-3}M

Now we have to calculate the rate of consumption.

\text{Rate of consumption}=\frac{\text{Concentration}}{\text{Time}}

\text{Rate of consumption}=\frac{1.2\times 10^{-3}M}{170s}

\text{Rate of consumption}=7.0\times 10^{-6}M/s

Thus, the rate of consumption of S_2O_2^{-8} is, 7.0\times 10^{-6}M/s

8 0
3 years ago
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