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:
The chlorine gas and potassium bromide solution react to form liquid bromine and potassium chloride solution.
Explanation:
Chemical equation:
Cl₂(g) + KBr (aq) → KCl (aq) + Br₂(l)
Balanced chemical equation:
Cl₂(g) + 2KBr (aq) → 2KCl (aq) + Br₂(l)
This equation showed that the chlorine gas and potassium bromide solution react to form liquid bromine and potassium chloride solution.
Chlorine is more reactive than bromine it displace the bromine from potassium and form potassium chloride solution.
The given equation is balanced and completely hold the law of conservation of mass.
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Answer:
d
Explanation:
because boron and fluorine are both nonmetals and don't fit the guidlelines for creating ionic compounds
If you cross a heterozygous organism with another heterozygous organism, pair two Tt and Tt. Below is a punnet square to show the result:
T t
T TT Tt
t Tt tt
This means that you have two heterozygous recessive and one homogenous dominant and one homogenous recessive.
