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3 years ago

CH3COOH mc015-1.jpg CH3COO– (aq) + H+(aq) What will happen to the chemical equilibrium of the solution if CH3COONa is added?

1 answer:

andrey2020 [161]3 years ago

7 0

The equilibrium will shift to the left or the backward reaction since addition of CH3COONa will add more CH3COO- ions to the solution. The formation of reactants are promoted.

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Answer:

in a solution of salt in water, the solute is salt, and solvent is water.

Explanation:

C) salt is the solute, water is the solvent.

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2 years ago

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3 years ago

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4 years ago

Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C

umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

$K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}$

Given:

$\left[I_2_{Equilibrium} \right]$ = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

$0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}$

So,

$\left[I_{Equilibrium} \right]^2=0.011\times 0.10$

$\left[I_{Equilibrium} \right]^2=0.0011$

$\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M$

Thus, The concentration of I at equilibrium = 3.3166×10⁻² M

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4 years ago

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution

ch4aika [34]

Answer: The molecular weight of protein is $1.14\times 10^2g/mol$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of protein = 400 mg = 0.4 g (Conversion factor: 1 g = 1000 mg)

$M_{solute}$ = molar mass of protein = ?

$V_{solution}$ = Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol$

Hence, the molecular weight of protein is $1.14\times 10^2g/mol$

4 0

3 years ago