1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules. Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas. The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.
2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride. So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required. Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
It would be 23, s choice C.
Complete Question
The complete question is shown on the first uploaded image
Answer:
Answer: 760 uM
Explanation:
the addition of solvent to a solution in a such away that the volume of the solution increases and the concentration decreases is a process know as dilution
The concentration and the volume of the dilute and concentrated solution are given below as follows
Concentrated solution Dilute solution
Given that M1 = 97.0uM
V1 = 47.0 mL
V2 = 6.0 mL
M2 = ?
Therefore M1V1 = M2V2
M2 = M1V1/V2
M2 = (97*47)/6
M2 = 760 uM
Answer: 760 uM
Answer:
0.0010m SO₄²⁻
Explanation:
The freezing point depression due the addition of a solute into a pure solvent follows the equation:
ΔT = Kf×m×i (1)
<em>Where ΔT are °C that freezing point decreases (273.15K - 272.47K = 0.68K = 0.68°C). Kf is the constant of freezing point depression (1.86°C/m), m is molality of the solution (0.1778m) and i is Van't Hoff factor.</em>
Van't Hoff factor could be understood as in how many one mole of the solute (sulfuric acid, H₂SO₄), is dissociated.
H₂SO₄ dissociates as follows:
H₂SO₄ → HSO₄⁻ + H⁺
HSO₄⁻ ⇄ SO₄²⁻ + H⁺
<em>Not all HSO₄⁻ dissociates.</em>
1 Mole of H₂SO₄ dissociates in 1 mole of H⁺+ 1 mole of HSO₄⁻ + X moles of SO₄²⁻= 2 + X
Replacing in (1):
0.68°C = 1.86°C/m×0.1778m×i
2.056 = i
Moles of SO₄²⁻ are 2.056 - 2 = 0.056moles SO₄²⁻.
If 1 mole has a concentration of 0.1778m, 0.056moles are:
0.056moles ₓ (0.1778m / 1mole) =
<h3>0.0010m SO₄²⁻</h3>
