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MAVERICK [17]
3 years ago
14

What is the solubility in moles/liter for silver chloride at 25 oC given a Ksp value of 1.6 x 10-10. Write using scientific nota

tion and use 1 or 2 decimal places (even though this is strictly incorrect!)
Chemistry
1 answer:
trapecia [35]3 years ago
5 0

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

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its proprionic acid which do not cause any hazardous effect on environment it only effect internal body of human if it is ingested.

Explanation:

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3 0
3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
A chemist burns 0.8 g of iron in oxygen.
stich3 [128]

Answer:

0.3g

Explanation:

1.1 - 0.8 = 0.3

Hope this helps

7 0
3 years ago
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