Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the soluti
on on the left. a. 0.035 M Ba(NO3)2; NaF b. 0.085 M Calz; K2SO4 c. 0.0018 M AgNO3; RbCI
1 answer:
Answer:
a) 0.0265M
b) 8.35×10^-4M
c) 9.833×10^-8M
Explanation:
First we must write down the equation of each reaction. After identifying the insoluble precipitate, we go ahead to write down the expression for its solubility product. If the solubility product of the compound is known, the concentration of the precipitating agent can now be calculated from the available information in the question.
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Answer:
21.6 g
Explanation:
The reaction that takes place is:
- CH₄ + 2O₂ → CO₂ + 2H₂O
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
☃️ Chemical formulae ➝
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
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