The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e
Answer:
Density = 4.191 gm/L
Explanation:
Given:
Molar mass = 93.89 g/mol
Volume(Missing) = 22.4 L (Approx)
Find:
Density at STP
Computation:
Density = Mass/Volume
Density = 93.89 / 22.4
Density = 4.191 gm/L
Answer:
umm.. B. a base that generates a lot of hydroxide ions in water.
Answer:
Shifts the equilibrium to the left. reduces solubility.
Explanation:
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S S 2S
∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²
⇒ 4S² * S = 6.4 E-9
⇒ 4S³ = 6.4 E-9
⇒ S³ = 1.6 E-9
⇒ S = 1.1696 E-3 M
- NaF(s) → Na+(aq) + F-(aq)
0.10M 0.10M 0.10M
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S' S' 2S' + 0.10
⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²
If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:
⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'
⇒ S' = 6.4 E-7 M
∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption
We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).
Answer : The enthalpy change during the reaction is -6.48 kJ/mole
Explanation :
First we have to calculate the heat gained by the reaction.
where,
q = heat gained = ?
m = mass of water = 100 g
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 23.4 kJ
n = number of moles barium chloride = 
Therefore, the enthalpy change during the reaction is -6.48 kJ/mole
