Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
I believe it’s C.) Mass. Hope I’m right.
Answer:
2.7005 g/cm²
explanation: as we know density=mass/volume
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
