I would say false hope that helped
Answer:
Explanation:
AgCl ⇄ Ag⁺ + Cl⁻
m m m
If x mole of AgCl be dissolved in one litre .
[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰
m² = 1.6 x 10⁻¹⁰
m = 1.26 x 10⁻⁵ moles
So solubility of AgCl is 1.26 x 10⁻⁵ moles / L
Answer:
The percent yield of this reaction is 84.8 % (option A is correct)
Explanation:
Step 1: Data given
The student isolated 15.6 grams of the product = the actual yield
She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams
Step 2: Calculate the percent yield
Percent yield = (actual yield / theoretical yield ) * 100 %
Percent yield = (15.6 grams / 18.4 grams ) * 100 %
Percent yield = 84.8 %
The percent yield of this reaction is 84.8 % (option A is correct)
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
#SPJ4
The answer is C: The plum-pudding model
