Answer:
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Explanation:
Mass percentage of oxalic acid = 5.50%
This means that in 100 grams of solution there are 5.50 grams of oxalic acid.
Mass of solution , m = 100
Volume of the solution = V
Density of the solution = d = 1.024 g/mL
V = 97.66 mL = 0.09766 L
(1 mL = 0.001 L)
Moles of oxalic acid = 
The molarity of the solution :
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
To determine what gas is this, we use Graham's Law of Effusion where it relates the rates of effusion of gases and their molar masses. We do as follows:
r1/r2 = √(M2 / M1)
Let 1 be the the unkown gas and 2 the H2 gas.
r1/r2 = 0.225
M2 = 2.02 g/mol
0.225 = √(2.02 / M1)
M1 = 39.90 g/mol
From the periodic table of elements, most likely, the gas is argon.
Its B addition
where an atom adds to the broken double bond of a hydrocarbon and saturates it
hope that helps
LiOH is going to neutralize the acid because it’s a base
Answer:
a) K = [ CO2(g) ]
⇒ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Explanation:
a) CaCO3(s) ↔ CaO(s) + CO2(g)
⇒ K = [ CO2(g) ]
∴ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) H2(g) + F2(g) ↔ 2 HF(g)
⇒ K = [ HF(g) ] ² / [ F2(g) ] * [ H2(g) ]
⇒ Kp = PHF² / PF2 * PH2
for ideal gas:
PV = RTn
⇒ P = n/V RT = [ ] RT
⇒ Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same.
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
