The oxidation numbers of nitrogen in NH4+, N2O5, and NaNO3 are, respectively:
2 answers:
Answer: Option (b) is the correct answer.
Explanation:
It is known that general charge on a hydrogen atom is +1, oxygen atom is -2, and sodium atom is +1.
Hence, oxidation number of nitrogen atom in the given species or compounds is as follows.
x + 4(1) = 1
x = 1 - 4
= -3
2x + 5(-2) = 0 (since total charge on the molecule is zero)
x = +5
1 + x + 3(-2) = 0
x = +5
Thus, we can conclude that the oxidation numbers of nitrogen in NH4+, N2O5, and NaNO3 are -3, +5, +5 respectively.
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Answer:
2.32
Explanation:
For the resolution of the problem one of the laws of Charles - Gay Lussac must be used, in which it is stated that if a<em> certain amount of gas is kept at a constant volume (as in this case), its pressure will be directly proportional to its absolute temperature.</em> Mathematically, it can be written as follows:
⇒to constant V
In the case of the statement, that we have an initial system (P₁ and T₁) and an final system (P₂ and T₂), the equation can be written:
It can be rearranged to clear P₂ which is what needs to be calculated, then:
It would only be necessary to <em>convert the temperature units from ° C to K</em>, to have the absolute temperature values. If T (K) = t (° C) + 273.15, then:
T₁ = 0 + 273.15 = 273.15 K
T₂ = 150 + 273.15 = 423.15 K
Now, you can replace the values to the equation and calculate P₂
≅ 2.32 atm
As the statement requests the result with three significant figures and no units, the answer is that<em> the final pressure is 2.32</em>