Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Isn't a chemical change like something that's not a physical change or physically changed but is something that uses natural chemicals? that's my guess sorry if it's wrong I think I'm wrong though
Answer:
Number of moles = 0.0005 mol.
Explanation:
Given data:
pH = 3
Volume of solution = 500 mL
Number of moles = ?
Solution:
HCl dissociate to gives H⁺ and Cl⁻
HCl → H⁺ + Cl⁻
It is known that,
pH = -log [H⁺]
3 = -log [H⁺]
[H⁺] = 10⁻³ M
[H⁺] = 0.001 M
Number of moles of HCl:
Molarity = number of moles / Volume in litter
Number of moles = Molarity × Volume in litter
Number of moles = 0.001 mol/L × 0.5 L
Number of moles = 0.0005 mol
Answer:
Hello
The answer is 18,7(b)
Explanation:we don't know the value of o² so at first we should put (x)instead of gr of o²and then write 1molO²/32grO²×2mol SO²/3mol O²×64gr SO²/1molSO²=25 gr SO².and then just find the value of (x).
Answer:
There are 0.5 mole in 20g of argon.
Explanation:
40 g of argon = 1mole
Then 20g of argon is,
→ 1/40 × 20
→ 0.5 mole
