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oee [108]
3 years ago
13

Determine the molar solubility of agi in pure water. ksp (agi) = 8.51 × 10-17.

Chemistry
1 answer:
Alex787 [66]3 years ago
3 0
When the salt AgI dissolves in water it dissociates as follows;
AgI ---> Ag⁺ + I⁻
molar solubility is the number of moles that are dissolved in 1 L of solution.
If molar solubility of AgI is x, then molar solubility of Ag⁺ is x and I⁻ is x.
the formula for solubility product constant - ksp of AgI is given below 
ksp = [Ag⁺][I⁻]
ksp = (x)(x)
ksp = 8.51 x 10⁻¹⁷ 
therefore,
x² = 8.51 x 10⁻¹⁷
x = 9.22 x 10⁻⁹
since molar solubility of AgI is x, then molar solubility of AgI is 9.22 x 10⁻⁹ M
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Natural gas burns in air to form carbon dioxide and water, releasing heat. CH4(g)+2O2(g)→CO2(g)+2H2O(g) ΔHrxn = -802.3 kJ.
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Answer:

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2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.

Explanation:

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Mass of water = m = 45.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 21.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

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Q = 3,950.1 J = 3.9501 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 3.950.1 kJ of heat will be given by:

=\frac{3.950.1 kJ}{802.3 kJ}=0.004923 mol

Mass of 0.004923 moles of methane :

0.004923 mol × 16 g/mol=0.0788 g

Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.

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Mass of water = m = 50.0 g

Specific heat capacity of water = c = 4.18 J/g°C

Change in temperature of water = ΔT = 26.0°C.

Heat required to raise the temperature of water by 21.0°C = Q

Q=mc\Delta T= 50.0 g\times 4.18 J/g^oC\times 26.0^oC

Q = 5,434 J= 5.434 kJ

According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.

Then 5.434 kJ of heat will be given by:

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6 0
4 years ago
A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O
Alexxandr [17]
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But as the yield is only 7.5% 

the required amount is    ((46/60) * 700 ) / 0.075 =  7155.56 g 

=  7.156 kg to nearest gram.  Answer




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Answer:

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