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erastovalidia [21]
3 years ago
9

Compartments A and B are separated by a membrane that is permeable to K+ but not to Na+ or Cl-. At time zero, a solution of KCl

is poured into compartment A and an equally concentrated solution of NaCl is poured into compartment B. Which would be true once equilibrium is reached?
Chemistry
1 answer:
nikklg [1K]3 years ago
5 0

Answer:

Incomplete question

Complete question:

8. Compartments A and B are separated by a membrane that is permeable to K+ but not to Na+ or Cl-. At time zero, a solution of KCl is poured into compartment A and an equally concentrated solution of NaCl is poured into compartment B. Which would be true once equilibrium is reached?

A. The concentration of Na+ in A will be higher than it was at time zero.

B. Diffusion of K+ from A to B will be greater than the diffusion of K+ from B to A.

C. There will be a potential difference across the membrane, with side B negative relative to side A.

D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.

E. The concentration of Cl- will be higher in B than it was at time zero.

Answer: D. The electrical and diffusion potentials for K+ will be equal in magnitude and opposite in direction.

Explanation:

Diffusion is the movement of molecules from region of higher concentration to lower concentration through a semipermeable membrane.

Since the k+ is the permeable membrane, the k+ ion in the KCl would move in equal magnitude and direction in the solution.

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Elina [12.6K]

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3 0
3 years ago
How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.
Slav-nsk [51]
Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0
3 years ago
Which of the following will have the highest boiling point?
DerKrebs [107]

Answer:

0.1M NH3

Explanation:

The boiling point of aqueous solutions depend on the nature of intermolecular interactions present. KBr will yield an ionic solution but NH3 will yield a molecular solution having hydrogen bonds. The degree of hydrogen bonding in the aqueous solution will further increase with the concentration of the solution.

Remember that experimental data shows that hydrogen bonds are strong bonds that lead to a significant increase in the boiling point of solutions. Hence 0.1M NH3 solution will have a higher boiling point due to intermolecular hydrogen bonding in the solution.

8 0
3 years ago
How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s
erastova [34]

Answer : The grams of carbon monoxide needed are 148.89 g

Solution : Given,

Mass of iron, Fe = 198.5 g

Molar mass of iron, Fe = 56 g/mole

Molar mass of carbon monoxide, CO = 28 g/mole

First we have to calculate the moles of iron, Fe.

\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{198.5g}{56g/mole}=3.545moles

The balanced chemical reaction is,

Fe_2O_3(s)+3CO(g)\rightarrow 3CO_2(g)+2Fe(s)

From the balanced reaction, we conclude that

2 moles of iron produces from the 3 moles of carbon monoxide

3.545 moles of iron produces from the \frac{3}{2}\times 3.545=5.3175 moles of carbon monoxide

Now we have to calculate the mass of carbon monoxide, CO.

\text{ Mass of CO}=\text{ Moles of CO}\times \text{ Molar mass of CO}

\text{ Mass of CO}=(5.3175moles)\times (28g/mole)=148.89g

Therefore, the grams of carbon monoxide needed are 148.89 g

7 0
3 years ago
5Br−+BrO3−+6H+→3Br2+3H2O
sashaice [31]

Explanation :

The balanced chemical reaction is,

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The expression for the rates of consumption of the reactants are:

The rate of consumption of Br^- = -\frac{1}{5}\frac{d[Br^-]}{dt}

The rate of consumption of BrO_3^- = -\frac{d[BrO_3^-]}{dt}

The rate of consumption of H^+ = \frac{1}{6}\frac{d[H^+]}{dt}

The expression for the rates of formation of the products are:

The rate of consumption of Br_2 = +\frac{1}{3}\frac{d[Br_2]}{dt}

The rate of consumption of H_2O = +\frac{1}{3}\frac{d[H_2O]}{dt}

5 0
3 years ago
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