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3 years ago

Use ksp=4.87×10−17 to calculate the solubility of iron (ii) hydroxide in pure water in grams per 100.0 ml of solution. g

1 answer:

5 0

The ksp = 4.87 × 10^-17
Therefore; 4.87 ×10^-17 = [Fe2+][OH-]^2 = (X)(2X)^2 = 4X^3
Hence;
4x^3 =4.87 × 10^-17
x = 2.30 × 10^-6 M
therefore the molarity is 2.3 ×10^-6 M
Thus;
The mass of Fe(OH)2 in 100 ml of water, will be given by:
= 2.30 × 10^-6 mol/L = 2.3 ×× 10^-5 mole/100 mL
= 89.86 g/mol × 2.3 × 10^-6 × 0.100 = 2.07 × 10^-5 g
= 2.07 × 10^-5 g
= 2.07 × 10^-5 g

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2-phosphoglycerate(2PG) is converted to phosphoenolpyruvate (PEP) by the enzyme enolase. The standard free energy change(deltaGo

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Answer:

The correct option is: (D) -2.4 kJ/mol

Explanation:

Chemical reaction involved: 2PG ↔ PEP

Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol

Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)

Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)

Reactant concentration: 2PG = 0.5 mM

Product concentration: PEP = 0.1 mM

Reaction quotient: $Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2$

To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:

$\Delta G = \Delta G^{\circ } + 2.303 R T log Q_{r}$

$\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)$

$\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)$

Therefore, the Gibb's free energy change at 37° C (310.15 K): ΔG = (-2.45 kJ/mol)

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3 years ago

The electrons in an insulator are tightly bound to their atoms.

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calculate the number of moles of sulfuric acid that is contained in 250 mL of 8.500 M sulfuric acid solution

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Answer : The moles of $H_2SO_4$ are, 2.125 mole.

Explanation : Given,

Molarity of $H_2SO_4$ = 8.500 M

Volume of solution = 250 mL = 0.250 L (1 L = 1000 mL)

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of }H_2SO_4}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$8.500M=\frac{\text{Moles of }H_2SO_4}{0.250L}$

$\text{Moles of }H_2SO_4=2.125mol$

Therefore, the moles of $H_2SO_4$ are, 2.125 mole.

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