Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
Answer:
Explanation:
We have to start with the <u>reaction</u>:

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:
C: 12 g/mol
O: 16 g/mol
Mg: 23.3 g/mol
If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:
In other words:
. With this in mind, we can calculate the moles:

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

With the molar mass of
(
. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:
I hope it helps!
Answer:
Acceleration due to gravity is reduced to half its value on the earth's surface at an altitude of 2.65×106 m
Explanation:
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