Antimony has two naturally occurring isotopes. Their abundance is given in the pic attached below
First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT
when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K
12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L
Answer:
2.2%
Explanation:
Percentage error,
You apply the formula,
[(Estimated value - Actual value)/Actual value] × 100%
; [(43.26 - 42.32)/42.32] × 100
; (0.94/42.32) × 100
; 0.022 × 100
Percent error = 2.2%
Answer:
13.4 (w/w)% of CaCl₂ in the mixture
Explanation:
All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.
To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.
<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>
0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻
<em>Moles CaCl₂:</em>
3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂
<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>
1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture
That means mass percent of CaCl₂ is:
0.207g CaCl₂ / 1.55g * 100 =
<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>