Mg(NO3)2 => <span>Magnesium nitrate
hope this helps!</span>
Answer: measure the mass (48.425g) of KCl
Explanation:
To prepare the solution 0.65M KCl we must measure the mass of KCl that would be dissolved in 1L of the solution. This can be achieved by:
Molar Mass of KCl = 39 + 35.5 = 74.5g/mol
Number of mole (n) = 0.65
Mass conc of KCl = n x molar Mass
Mass conc of KCl = 0.65 x 74.5 = 48.425g
Therefore, to make 0.65M KCl, we must measure 48.425g
Answer:
Oxidation number of F2O = 0−(−1×2)
State of oxygen will be=+2
Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble the data in one place.
2Mg + O₂ ⟶ 2MgO
n/mol: 2 5
Calculate the moles of MgO we can obtain from each reactant.
From Mg:
The molar ratio of MgO:Mg is 2:2
From O₂:
The molar ratio of MgO:O₂ is 2:1.
Answer:
Explanation:
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
ionisation constant = 1.36 x 10⁻⁴ .
molecular weight of lactic acid = 90 g
moles of acid used = 20 / 90
= .2222
it is dissolved in one litre so molar concentration of lactic acid formed
C = .2222M
Let n be the fraction of moles ionised
CH₃CHOHCOOH ⇄ CH₃CHOHCOO⁻ + H⁺
C - nC nC nC
By definition of ionisation constant Ka
Ka = nC x nC / C - nC
= n²C ( neglecting n in the denominator )
n² x .2222 = 1.36 x 10⁻⁴
n = 2.47 x 10⁻²
nC = 2.47 x 10⁻² x .2222
= 5.5 x 10⁻³
So concentration of hydrogen or hydronium ion = 5.5 x 10⁻³ g ion per litre .
