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ra1l [238]
3 years ago
13

What is the molar solubility of barium fluoride ( baf2 ) in water? the solubility-product constant for baf2 is 1.7 × 10-6 at 25∘

c?
Chemistry
1 answer:
aliina [53]3 years ago
3 0
Molar solubility is the number of moles that are dissolved in 1 L solution. 
when BaF₂ dissolves it dissociates into the following ions
BaF₂ --> Ba²⁺ + 2F⁻
if the molar solubility of BaF₂ is X, then molar solubility of Ba²⁺ is X and F⁻ is 2x
then the formula for the solubility product constant -ksp is;
ksp = [Ba²⁺][F⁻]²
ksp = X * (2X)²
ksp = 4X³
since ksp = 1.7 x 10⁻⁶
4X³ = 1.7 x 10⁻⁶
X = 0.0075 M
molar solubility of BaF₂ is 0.0075 M
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MAGNESIUM (Mg)<br> +FLUORINE (F) =
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Answer:

MgF2 magnesium fluoride

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3 years ago
Which option explains how some parts of a plant can be white?
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Note: This question is incomplete because options are missing. Nevertheless, I can answer it based on my prior knowledge.

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7 0
3 years ago
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noname [10]

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8 0
3 years ago
Calculate the percent yield when 500 grams of carbon dioxide react with an excess of water to produce 640 grams of carbonic acid
Elena-2011 [213]

Answer:

Percent yield =  90.5%

Explanation:

Given data:

Mass of carbon dioxide = 500 g

Mass of water = excess

Actual yield of carbonic acid = 640 g

Percent yield = ?

Solution:

Balanced chemical equation:

CO₂ + H₂O  → H₂CO₃

Number of moles of carbon dioxide

Number of moles  = Mass / molar mass

Number of moles = 500 g/ 44 g/mol

Number of moles = 11.4 mol

Now we will compare the moles of H₂CO₃ with CO₂.

                              CO₂          :              H₂CO₃

                                 1             :                  1

                               11.4           :                11.4

Mass of carbonic acid:

Mass = number of moles × molar mass

Mass = 11.4 mol × 62.03 g/mol

Mass = 707.14 g

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Percent yield = actual yield / theoretical yield × 100

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7 0
4 years ago
A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
givi [52]

Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

V_{V^{+3}} = 59.0 \ mL

V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
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