Answer:
1. Volume as STP = 755 L
2. Outside temperature = 255 K
3. Percentage yield = 70.5%
Explanation:
1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K
Using the general gas equation :
P1V1/T1 = P2V2/T2
P1 = 620 kpa
V1 = 140 L
T1 = 37°C or (273.15 + 37) K = 310.15 K
P2 = 101.3 kpa
V2 = ?
T2 = 273.15 K
V2 = P1V1T2/P2T1
V2 = 620 × 140 × 273.15 / 101.3 × 310.15
V2 = 755 L
2. Using Charles' gas law:
V1/T1 = V2/T2
V1 = 2.5 L
T1 = 290 K
V2 = 2.2 L
T2 = ?
T2 = V2T1/VI
T2 = 2.2 × 290 / 2.5
T2 = 255 K
3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu
From equation of the reaction, 2 moles of Al produces 3 moles of Cu
Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g
2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g
54 g of Al produces 190.5 g of Cu
1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu
Percentage yield = actual yield /theoretical yield × 100%
Percentage yield = 4.65/6.60 × 100%
Percentage yield = 70.5%
Answer:
ΔH°f P4O10(s) = - 3115.795 KJ/mol
Explanation:
- P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
- ΔH°rxn = ∑νiΔH°fi
∴ ΔH°rxn = - 327.2 KJ
∴ ΔH°f H2O(l) = - 285.84 KJ/mol
∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol
⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ
⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2
⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol
Protons have a positive charge.
How I remember is Protons and Positive both start with the letter P.
Answer:
Molarity of the sodium hydroxide solution is 1.443 M/L
Explanation:
Given;
0.60 M concentration of NaOH contains 2.0 L
3.0 M concentration of NaOH contains 495 mL
Molarity is given as concentration of the solute per liters of the solvent.
If the volumes of the two solutions are additive, then;
the total volume of NaOH = 2 L + 0.495 L = 2.495 L
the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M
Molarity of NaOH solution = 3.6 / 2.495
Molarity of NaOH solution = 1.443 M/L
Therefore, molarity of the sodium hydroxide solution is 1.443 M/L
Answer:
first of all is that really chemistry
