Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
Answer:
Yes, water can stay liquid below zero degrees Celsius. There are a few ways in which this can happen. The freezing point of water drops below zero degrees Celsius as you apply pressure. When we apply pressure to a liquid, we force the molecules to get closer together.
Explanation:
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Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of phosphorus = 25.0 g
Mass of oxygen = 50.0 g
What is limiting reactant ?
Solution:
Chemical equation:
P₄ + 5O₂ → P₄O₁₀
Number of moles of P₄:
Number of moles = mass/molar mass
Number of moles = 25.0 g/ 123.89 g/mol
Number of moles = 0.20 mol
Number of moles of O₂:
Number of moles = mass/molar mass
Number of moles = 50.0 g/ 32 g/mol
Number of moles = 1.56 mol
now we will compare the moles of reactants with product:
P₄ : P₄O₁₀
1 : 1
0.20 : 0.20
O₂ : P₄O₁₀
5 : 1
1.56 : 1/5×1.56 = 0.312 mol
Less number of moles of product are formed by the oxygen thus it will act as limiting reactant.
Answer:
Nhiệt dung riêng của đồng là
380
J
/
k
g
.
K
Explanation:
