<span>4: Form An Aqueous Solution
This is the only answer that can be observed without testing gear and with the naked eye.... Hope I helped ^-^</span>
Answer:
5.158 mol/L
Explanation:
To find the molarity, you need to use the formula:
Molarity (M) = moles / volume (L)
You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.
<u>Steps 1 - 2:</u>
2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂
6.5287 g Na₂CO₃ 1 mole 2 moles HCl
-------------------------- x ------------- x ------------------------- = 0.12318 mole HCl
106 g 1 mole Na₂CO₃
<u>Step 3:</u>
23.88 mL / 1,000 = 0.02388 L
<u>Step 4:</u>
Molarity = moles / volume
Molarity = 0.12318 mole / 0.02388 L
Molarity = 5.158 mole/L
**mole/L is equal to M**
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
Answer:
the answer is D
Explanation:
percentage composition= mole of the substance divided by the total molar mass of the compound multiplied by 100.
