<span>d.2HNO3 (aq) + Sr(OH)2 (aq) → 2H2O (l) + Sr(NO3)2(aq)
4H </span>4H
8O 8O
2N 2N
1Sr 1Sr<span>
</span>
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
Balance Chemical Equation for combustion of Propane is as follow,
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
According to equation,
1 mole of C₃H₈ on combustion gives = 4 moles of H₂O
So,
5 moles of C₃H₈ on combustion will give = X moles of H₂O
Solving for X,
X = (5 mol × 4 mol) ÷ 1 mole
X = 20 moles of H₂O
Calculating number of molecules for 20 moles of H₂O,
As,
1 mole of H₂O contains = 6.022 × 10²³ molecules
So,
20 moles of H₂O will contain = X molecules
Solving for X,
X = (20 mole × 6.022 × 10²³ molecules) ÷ 1 mol
X = 1.20 ×10²⁵ Molecules of H₂O
Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.
Explanation:
Given:
= 2.25 L,
= 9.0 mol
= 1.85 L,
= ?
Formula used to calculate the moles of helium are as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.
Answer:
Explanation:
<em>0.5 i go to k12 i jus took the test</em>