Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
Electrons are the thing in the outer energy levels
Answer: D) Eight
Just pretend this part doesn't exist nope nothing to see here the answer is correct on edge2020 I swear on my brainly points
Avagadro's law gives the relationship between volume of gas and amount of moles of gas. It states that at constant temperature and pressure, volume of gas (V) is directly proportional to number of moles of gas (n).
V/n = k
where k - constant
V1 = 42.0 L
n1 = 1.90 mol
n2 = 1.90 mol - 0.600 = 1.30 mol
substituting the values in the equation
V = 28.7 L
Volume of the gas is 28.7 L
Answer:
Explanation:
Standard enthalpy of combustion (ΔH∘C) is the enthalpy change when 1 mole of a substance burns under standard state conditions;
