Answer:
The number of moles of potassium hydroxide, KOH required to make 4 moles of K₂SO₄ is 8 moles of KOH
Explanation:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
From the above reaction, we have 2 moles of KOH combining with 1 mole of H₂SO₄ to produce 1 mole of K₂SO₄ and 2 moles of H₂O.
Therefore the number of moles of potassium hydroxide that will be needed to make 4 moles of K₂SO₄ is;
8KOH + 4H₂SO₄ → 4K₂SO₄ + 8H₂O
8 moles of KOH is required to make 4 moles of K₂SO₄.
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
The tertiary structure of a protein includes all of the following interactions except p<span>eptide</span> bonds.
-Hope this helps.
Answer:
=> 572.83 K (299.83°C).
=> 95.86 m^2.
Explanation:
Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.
Hence;
Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.
15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).
If we solve for x, we will get the value of x to be;
x = 572.83 K (2.99.83°C).
x is the temperature of the exit gas that is 572.83 K(299.83°C).
(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).
∆T = 93.87/ln 1.38.
∆T = 291.521K.
Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.
heat-transfer area = 1931146.394/69.1 × 291.521.
heat-transfer area= 95.86 m^2.
