The given question is incomplete. The complete question is:
How much heat is produced when 24.8 g of
is burned in excess oxygen gas
Given:
ΔH= −802 kJ.
Answer: 1243.1 kJ
Explanation:
Heat of combustion is the amount of heat released on complete combustion of 1 mole of substance.
Given :
Amount of heat released on combustion of 1 mole of methane = 802 kJ kJ/mol
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number
of particles.
1 mole of
weighs = 16 g
Thus we can say:
16 g of
on combustion releases heat = 802 kJ
Thus 24.8 g of
on combustion releases =
Thus heat released when 24.8 g of methane is burned in excess oxygen gas is 1243.1 kJ
Answer:-
(a) 3.5
(b) 3
Explanation:-
2KClO3 --> 2KCl + 3 O2
From the equation we see that 2 moles of KClO3 gives 2 moles of KCl.
So 3.5 moles of KClO3 will give 3 moles of KCl.
Again
3 moles of O2 are produced with 2 moles of KCl.
If 4.5 moles of O2 produced then
moles of KCl = 4.5 x 2/3
=3
<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
The molecules are frozen in place but still vibrate
Answer:
I think <em><u>alpha</u></em> and <em><u>beta</u></em> is the answer.