The net equations are obtained from the double displacement of the cations and anions, then balance.
NH3(aq) + HC2H3O2 (aq) = NH4+(aq) + C2H3O2-(aq<span>)
</span><span>H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq)</span><span>
</span><span>2NaOH(aq) + H2SO4 (aq) = Na2SO4 (s)+ 2H2O (aq)
</span>H2S (aq) + Ba(OH)2 (aq) = BaS (s)+ 2H2O (aq)
The correct answer is 1 amu; neutral
Answer:
See Explanation Below
Explanation:
A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.
B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5
C) The rate will increase by 4 since you are doubling both you have to multiply them both.
Answer:
398 mL
Explanation:
Using the equation for molarity,
C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L
V₂ = V₁ + V' where V' = volume of water added.
So, From C₁V₁ = C₂V₂
V₁ = C₂V₂/C₁
= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L
= 0.875 mol/8.61 mol/L
= 0.102 L
So, V₂ = V₁ + V'
0.5 L = 0.102 L + V'
V' = 0.5 L - 0.102 L
= 0.398 L
= 398 mL
So, we need to add 398 mL of water to the nitric solution.