Answer:
= 200 mL
Explanation:
Using the dilution formula;
M1V1 = M2V2 ;
Where, M1 is the concentration before dilution, V2 is the volume before dilution, while M2 is the concentration after dilution and V2 is the volume after dilution.
M1 = 2.0 M
V1 = 50 mL
M2 = 0.50 M
V2 = ?
V2 = M1V1/M2
= ( 2.0 × 50 )/ 0.5
= 200 mL
Therefore, the volume after dilution will be, 200 mL
False.
False.
True.
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The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.
b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.
c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.
<span>Molten barium
chloride is separetes:</span><span>
BaCl</span>₂(l) →
Ba(l) + Cl₂(g), <span>
but first ionic bonds in this salt are separeted
because of heat:
BaCl</span>₂(l) →
Ba²⁺(l) + 2Cl⁻(l).
Reaction of reduction
at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation
at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
The anode is positive
and the cathode is negative.
A. Of a given element have different numbers of neutrons
