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3 years ago

10.222 g sample of hydrated barium iodide (Bal) is heated to dry off the water of

Chemistry

1 answer:

stealth61 [152]3 years ago

4 0

Answer:

BaI₂.5H₂O

Explanation:

Given Data:

Mass of Hydrated BaI₂ = 10.222 g

Mass of dried BaI₂ = 9.520 g

Mass of Water removed = 10.222-9.520 = 0.702 g

M.Mass of BaI₂ = 391.136 g/mol

M.Mass of Water = 18.02 g/mol

Now,

Calculate moles of dried BaI₂ as,

Moles = Mass / M.Mass

Moles = 9.520 g / 391.136 g/mol

Moles = 0.02434 moles

Calculate moles of Water as,

Moles = Mass / M.Mass

Moles = 0.702 g / 18.02 g/mol

Moles = 0.0389 moles

Then,

Calculate Mole ratio of BaI₂ and water as,

= 0.02434 moles BaI₂ / 0.0389 moles Water

= 0.625

Now,

We will convert this mole ratio to a whole number by multiplying it with a nearest integer,

= 0.625 × 8

= 5

Hence, this means for every one mole of BaI there are 5 moles of Water.

Result:

BaI₂.5H₂O

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How do you Convert 19mL to pL?

s2008m [1.1K]

Picoliter is a unit of measurement for liquids.

One picoliter = 1×10⁻⁹ mililiters

So:

19 mL ---- x pL

1×10⁻⁹ mL ---- 1 pL

1×10⁻⁹x = 19

x = 1,9 × 10¹⁰ pL

or 19,000,000,000 pL

Answer: 1.9 × 10¹⁰ pL

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1 year ago

Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in

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<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
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3 years ago

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Answer:

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Explanation:

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