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OlgaM077 [116]
3 years ago
7

KMnO4 + H2 S o4 + H2 S o4 is equal to K 2 S o4 + cuso4 + H2O + CO2​

Chemistry
1 answer:
Eddi Din [679]3 years ago
6 0
Do you need us to balance it?
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Inessa [10]

E

It does all those things

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3 years ago
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What is the charge on an ion that has an atomic number of 16 and contains 14e-?
lora16 [44]
The answer would be 2+ since the atomic number represents how many protons are in the element. In this case, there are 16 protons, but only 14 electrons, which means there are an additional 2 protons, hence the 2+ charge on the ion.
3 0
2 years ago
Write the balanced nuclear equation for β− decay of silicon−32. Include both the mass numbers and the atomic numbers with each n
weqwewe [10]

Answer:

^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e

Explanation:

Beta decay conserves the lepton number. In \beta^- decay, the atomic number of the element increases which is accompanied by the release of  an electron antineutrino, e^-+\bar{v}_e.

For example:-

^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e

The \beta^- decay of silicon-32 is shown below as:-

^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e

8 0
3 years ago
For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0
3 years ago
Newtons second law in your own words.
Misha Larkins [42]

Answer:

behavior of objects for which all existing forces are not balanced.

Explanation:

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3 years ago
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