Answer:
4
Explanation:
there is a decimal place present. so you would take away any zeros before the number.
Answer:
Hello
The answer is 25gr H2O
Explanation:
50grH2×1 molH2/2 gr H2×2mol H2O/2molH2=25gr H2O
Answer:
Explanation:
NaOH + HNO₃ ⟶ NaNO₃ + H₂O
There are two energy flows in this reaction.
Data:
V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹
V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹
T₁ = 35.00 °C; T₂ = 37.00 °C
Calculations:
(a) q₁
We have equimolar amounts of NaOH and HNO₃
n = 0.0300 mol
q₁ = 0.0300ΔH
(b) q₂
V = 100.0 mL + 100.0 mL = 200.0 mL
m = 200.0 g
ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C
q₂ = 200.0 × 4.184 × 2.00 = 1674 J
(c) ΔH
0.0300ΔH + 1674 = 0
0.0300ΔH = -1674
ΔH = -1674/0.0300
ΔH = -55 800 J/mol
ΔH = -55.8 kJ/mol
Answer:
D) 1.6 x 10⁻¹⁴
Explanation:
The solubility-product equilibrium constant for Ag₂SO₃ is given by the expression
Ksp = [Ag⁺]² [SO₃²⁻]
where [Ag⁺] and [SO₃²⁻] are the concentration of the species dissolved in solution for the equlibrium
Ag₂SO₃ (s) ⇄ 2 Ag⁺ + SO₃²⁻
we are given the concentration of Ag⁺ and from the stoichiometry of the equilibrium, the concentration of SO₃²⁻ is half that value, so
[Ag⁺]² = 3.2 x 10⁻⁵ M
[SO₃²⁻] = 3.2 x 10⁻⁵ M / 2 = 1.6 x 10⁻⁵ M
plugging these values into the solubility product constant equation we have
Ksp = (3.2 x 10⁻⁵)² x (1.6 x 10⁻⁵) = 1.6 x 10¹⁴
Therefore D is the correct answer.
Formula for Barium Nitrate = Ba(NO3)2
Thus based on stoichiometry:
1 mole of Ba(NO3)2 contains 2 moles of NO3-
Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M
Use the relation:
V1M1 = V2M2
V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L
Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.
