Answer:
Percent yield of PI3 = 95.4%
Explanation:
This is the reaction:
2P (s) + 3I2 (g) > 2PI3 (g)
Let's determine the moles of iodine that has reacted.
58.6 g / 253.8 g/mol = 0.231 mol
Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.
3 moles of I2 react to make 2 moles of PI3
0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3
As we have produced 0.147 moles let's determine the percent yield.
(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%
Answer:
12.29 M
Explanation:
- The reaction that takes place is:
H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O
- Now let's calculate the <u>moles of H₂SO₄ that were titrated</u>:
= 0.01229 mol H₂SO₄.
- Thus, the <u>concentration of the diluted solution is</u>:
0.01229 mol H₂SO₄ / 0.010 L = 1.229 M
- Finally, the <u>concentration of the original acid solution is:</u>
= 12.29 M
Answer:
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)
Explanation:
If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.
The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.
On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.
your answer will be :
B. <u>Na has a lower</u> <u>electronegativity than H</u>
because Na belongs to alkali metals which are least electronegative (most electro positive) but hydrogen is a non metal, it has higher electronegativity as compared to metals like Sodium (Na).