How many moles would be in 85.OmL of 0.750M KOH?
1 answer:
Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
Explanation:
Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L
Molarity = 0.750 M
It is known that molarity is the number of moles of solute present in liter of a solution.
Therefore, moles present in given solution are calculated as follows.
Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
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The missing data in the question is
The equilibrium constant (K) of the reaction below is K = 6.0 x 10⁻² with initial concentrations as follows: H₂ = 1.0 x 10⁻² M, N₂ = 4.0 M, and NH₃ = 1.0 x 10⁻⁴M.
<h3>What is Chemical Equilibrium?</h3>When the rate of forward reaction is equal to the rate of the backward reaction then the equation is said to be in Chemical Equilibrium.
N₂ + 3H₂ --> 2NH₃
The reaction quotient, Q, has the same algebraic expressions but uses the actual concentrations of reactants.
[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
Q = (5.6 x 10⁻³ )² / (1.0 x 10⁻²) ³(4.0 )
Q = 7.84
As Q > K,
The chemical system will shift to the left in order to produce more N₂ , H₂
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