Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
Answer:
Explanation:
k stand for equilibrium constants in terms of reaction
The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.
Consider the example below:
⇄
where
For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.
Answer: 4.5 x 10e-7
Explanation: 450 x 1e+9 = correct answer
Multiply amount of nanometers by 1e+9 to get the approximate result in meters.
<h3>
Answer:</h3>
1.93 g
<h3>
Explanation:</h3>
<u>We are given;</u>
The chemical equation;
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ
We are required to calculate the mass of ethane that would produce 100 kJ of heat.
- 2 moles of ethane burns to produce 3120 Kilo joules of heat
Number of moles that will produce 100 kJ will be;
= (2 × 100 kJ) ÷ 3120 kJ)
= 0.0641 moles
- But, molar mass of ethane is 30.07 g/mol
Therefore;
Mass of ethane = 0.0641 moles × 30.07 g/mol
= 1.927 g
= 1.93 g
Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g
Ok the ML (the 3rd number) is not legit because the ML value can only be from -L to L (the second value)
