Answer:
0.31 mol K x 39.1 grams K/ 1 mol k = 12.1 g
Explanation:
molar mass of K: 39.1
In order to find the mass in 0.31 mol of K, you must covert moles to grams. In one mole of Potassium there is 39.1 grams.
<span>284 g
First, lookup the atomic weights of all the elements involved.
Atomic weight of Calcium = 40.078
Atomic weight of Chlorine = 35.453
Now calculate the molar mass of CaCl2
40.078 + 2 * 35.453 = 110.984
Using that molar mass, calculate how many moles of CaCl2 you have.
445 g / 110.984 g/mol = 4.009586967 mol
Since each molecule of CaCl2 has 2 chlorine atoms, multiply the number of moles of CaCl2 by 2 to get the number of moles of Chlorine atoms.
4.009586967 * 2 = 8.019173935
And finally, multiply by the atomic weight of chlorine.
8.019173935 * 35.453 = 284.3037735
Since you have have 3 significant figures in your data, round the result to 3 significant figures, giving 284 grams.</span>
Answer:
1.008moles of iodine
Explanation:
Hello,
This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.
Percent yield = (actual yield / estimated yield) × 100
Actual yield = 1.2moles
Estimated yield = ?
Percentage yield = 84%
84 / 100 = 1.2 / x
Cross multiply and solve for x
100x = 84 × 1.2
100x = 100.8
x = 100.8/100
x = 1.008moles
1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI
Answer is: the solubility of silver oxalate is <span>a. 1.4 × 10-4 m.
</span>
<span>Chemical reaction
(dissociation) of silver oxalate in water:
Ag</span>₂C₂O₄(s) → 2Ag⁺(aq) + C₂O₄²⁻<span>(aq).
Ksp(Ag</span>₂C₂O₄) = [Ag⁺]²·[C₂O₄²⁻<span>].
[C</span>₂O₄²⁻] = x; solubility of oxalate ion.
[Ag⁺] = 2[C₂O₄²⁻<span>] = 2x
1.0·10</span>⁻¹¹ = (2x)² · x = 4x³.
x = ∛1.0·10⁻¹¹ ÷ 4.
x = 1.4·10⁻⁴ M.
