2KOH + Cu(NO3)2 → 2KNO3 + Cu(OH)2
2K⁺ 1Cu²⁺ 2K⁺ 1Cu²⁺
2OH ⁻ 2NO3⁻ 2NO3⁻ 2OH⁻
Displaced volume :
<span> Vf - Vi
</span>
77.0 mL - 50.0 mL => 27 mL
Therefore:
d = m / V
d = 521 / 27
d = 19.29 g/mL
Na2C2O4(aq) + CaCl2(aq) -----> 2NaCl(aq) + CaC2O4(s)
Here, CaC2O4(s) is a precipitate in the reaction as a result of precipitation reaction or double displacement reaction.
As we know that double displacement reaction two metal ions displaces each other from their salt solutions.
As we know that precipitation reaction is a reaction in which precipitate is formed.
Answer:
Explanation:
2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)
Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law
P . V = n . R .T
2.19 atm . 6.82L = n . 0.082 . 308K
(2.19 atm . 6.82L) / (0.082 . 308K) = n
0.591 mol = n
1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole
0.591 .2 = 1.182 mole of acid.
Molar mass of HCl = 36.45 g/m
1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g
Density HCl = HCl mass / HCl volume
0,118 g/mL = 43.1 g / HCl volume
43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)