From the equation:
4mol Li react with 1 mol O2
Molar mass Li = 7g/mol
mol in 84g Li = 84/7 = 12 mol Li
From the equation - 12 mol Li will react with 3 mol O2
At STP 1 mol O2 has volume = 22.4L
<span>
At STP 3 mol O2 has volume = 3*22.4 = 67.2L O2 gas will react. </span>
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
Answer:
Part A
The volume of the gaseous product is
Part B
The volume of the the engine’s gaseous exhaust is
Explanation:
Part A
From the question we are told that
The temperature is
The pressure is
The of
The chemical equation for this combustion is
The number of moles of that reacted is mathematically represented as
The molar mass of is constant value which is
So
The gaseous product in the reaction is and water vapour
Now from the reaction
2 moles of will react with 25 moles of to give (16 + 18) moles of and
So
1 mole of will react with 12.5 moles of to give 17 moles of and
This implies that
0.8754 moles of will react with (12.5 * 0.8754 ) moles of to give (17 * 0.8754) of and
So the no of moles of gaseous product is
From the ideal gas law
making V the subject
Where R is the gas constant with a value
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is
The volume is
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
Substituting values
heres your answer mate...
Answer: <u><em>Hope this helps</em></u>
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Explanation:</h3><h3>
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