Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
Explanation:
2H2 + O2 = 2H2O
2mol. 1mol. 2mol
2mol reacts with 1mol
13mol reacts with x
x=<u>13mol</u><u> </u><u>×</u><u> </u><u>1mol</u>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2mol</u>
x= <u>13mol</u>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>2mol
x= 6.5mol of oxygen
Answer:
1. smaller. 2. smaller. 3. greater
Explanation:
1. H−O−H angle is 104.45 and H−C−H angle is 109.5
2. O−S−O angle is 119 and F−B−F angle is 120
3. The F−S−F bond angle in SF₆ is 90 and F−Br−F bond angle in BrF₅ is 84.8
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
Mass defect Δ m = 236.052595 - 235.854000 = 0.198 amu
Reaction energy released Q = Δ m * 931.5
= 0.198 * 931.5 = 185 MeV
As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.
Solution for problem:
When 1 mole Neon (Ne) occupies 22.4 dm³ at STP then the volume occupied by 2.25 moles of Neon is calculated as,
= ( 22.4 dm³ × 2.25 moles ) ÷ 1 mole
= 50.4 dm³ 1dm³ = 1 L
Result:
So, 50.4 dm³ (Liter) volume will be occupied by 2.25 moles of Neon gas if it acts ideally at STP.
