Answer:
Molarity = 0.95 mol/dm³
Explanation:
Given data:
Volume of H₂SO₄ = 15.8 cm³
Volume of NaOH = 20 cm³
Concentration of NaOH = 1.5 mol/dm³
Concentration of H₂SO₄ = ?
Solution:
Chemical equation:
NaOH + H₂SO₄ → Na₂SO₄ + H₂O
First of all we will calculate the number of moles of NaOH and for that we will convert the units first,
Volume = 20 cm³/1000 = 0.02 L
Concentration of NaOH = 1.5 mol/dm³
1 mol/dm³ = 1 mol/L
Concentration of NaOH = 1.5 mol/L
Number of moles of NaOH:
Molarity = number of moles / volume in L
1.5 M = number of moles / 0.02 L
Number of moles = 1.5 M ×0.02 L
Number of moles = 0.03 mol
Now we will compare the moles of NaOH and H₂SO₄
NaOH : H₂SO₄
2 : 1
0.03 : 1/2×0.03 = 0.015 mol
Concentration of H₂SO₄:
Volume of H₂SO₄:
15.8 cm³/1000 = 0.0158 L
Molarity = number of moles / volume in L
Molarity = 0.015 mol / 0.0158 L
Molarity = 0.95 mol/L
1 mol/L = 1 mol/dm³
Molarity = 0.95 mol/dm³
Answer:−329.4 calorie
Explanation:Heat released by metal = q=m_mc_m(T_2-T1)q=m
m
c
m
(T
2
−T1)
m_m=50\ g\\c_m=0.108\ cal\ g^{-1}\ \degree C^{-1}\\q=50\times0.108(17-78)=-329.4\ caloriem
m
=50 g
c
m
=0.108 cal g
−1
°C
−1
q=50×0.108(17−78)=−329.4 calorie
Here minus sign indicate that heat is released
Answer:
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Answer:
2KClO3 ==> 2 KCl + 3O2
393g O2 x 1 mole/32 g = 12.28 moles O2
moles KClO3 needed = 12.28 mol O2 x 2 mol KClO3/3mol O2 = 8.19 mol
grams KClO3 = 8.19 moles x 123g/mol = 1007g
Explanation:
