It is a weighted average of the atomic masses of the naturally occurring isotopes of the element.
Answer: Hence, 2 moles of NaOH would neutralize 1 mole of 
Explanation:
One mole of hydrochloric acid neutralizes one mole of sodium hydroxide to give one mole of sodium chloride and one mole of water
When sulfuric acid is used in the place of HCl , 1 mole of sulfuric acid will neutralizes the 2 mole of sodium hydroxide and gives one mole of sodium sulfate and 2 moles of water. As we can see from the reaction:
Hence, 2 moles of NaOH would neutralize 1 mole of
This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions
Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
The excess base remaining is - 0.018 - 0.013 = 0.005 mol
total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL
[OH⁻] = 0.005 mol/0.075 L = 0.067 M
pOH = -log[OH⁻]
pOH = -log(0.067 M)
pOH = 1.17
pOH + pH = 14
Therefore pH = 14 - 1.17 = 12.83
by knowing pH we can calculate the [H₃O⁺]
pH = -log [H₃O⁺]
[H₃O⁺] = antilog[-12.83]
[H₃O⁺]= 1.47 x 10⁻¹³ M
Answer:
We wouldn't have Coronavirus
Explanation:
Answer:
80.33 %.
Explanation:
<em>Mg(s) + 2HCl(aq) → MgCl₂ + H₂.</em>
- Every 1.0 mole of Mg is dissolved in 2.0 moles of HCl and produce 1.0 mole of MgCl₂ and 1.0 mol H₂.
- To get the theoretical yield of hydrogen gas (in L):
- We want to calculate the no. of moles of Mg in 1.50 g:
n = mass/atomic mass = (1.50 g)/(24.3 g/mol) = 0.062 mol.
<u><em>Using cross multiplication:</em></u>
1.0 mol of Mg produces → 1.0 mol of hydrogen gas.
0.062 mol of Mg produces → 0.062 mol of hydrogen gas.
∵ PV = nRT.
<em>∴ V of hydrogen (the theoretical yield) = nRT/P </em>= (0.062 mol)(0.082 L.atm/mol.K)(295.0 K)/(1.0 atm) = <em>1.50 L.</em>
The actual yield of the reaction = 1.205 L.
<em>∴ The percent yield of the reaction = (actual yield)/(theoretical yield) x 100</em> = (1.205 L)/(1.50 L) x 100 = <em>80.33 %.</em>
