Answer:
22.9 Liters CO(g) needed
Explanation:
2CO(g) + O₂(g) => 2CO₂(g)
? Liters 32.65g
= 32.65g/32g/mol
= 1.02 moles O₂
Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)
∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)
Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.
∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed
___________________
*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).
For 2 it’s 2ft 10in
For 5 it’s 3ft 6in
You: put how tall you are
Adult: you can either measure an adult or use this 5 ft 10in
Potassium Cyanide, is the compound of KCN.
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
