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PolarNik [594]
3 years ago
14

I NEED HELP!! When cells are submerged in a hypertonic solution, the salt concentration inside the cells is _______ than the sal

t concentration outside the cells. This causes water to _______ the cells.
Chemistry
1 answer:
jekas [21]3 years ago
7 0
"Hypotonic" is the first one, and "Rush" is your second one.
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Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and oxalic acid (H2C2
belka [17]

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Oxalic acid is a dicarboxylic acid and forms sodium salt with NaOH and water

Explanation:

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A gas mixture with a total pressure of 765 mmHg contains each of the following gases at the indicated partial pressures: 134 mmH
ladessa [460]

Answer: c

Explanation:

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Lipid vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high con
NNADVOKAT [17]

Answer:

The question is not complete, the complete question should be "Lipids vesicles are formed containing pure water. If these vesicles are transferred to a solution that contains a rather high concentration of solutes, the solution outside the vesicle is said to be Hypertonic. True or False"

The answer is True

Explanation:

This is because  it contains greater concentration of solutes on the outside of the cell than the increase.

In other words hypertonic solutions have more concentrate of solutions on the outside than the inside.

3 0
3 years ago
In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
Mamont248 [21]

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

5 0
3 years ago
It is found when an unknown solid solute is dissolved in water that the solubility of the solid increases if temperature is decr
mina [271]

Answer:

exothermic entropy is increased

Explanation:

An exothermic process is one whose rate increases when the temperature is decreased. Hence if a decrease in temperature favours the dissolution of more solute at equilibrium, then the process is exothermic.

Similarly, the dissolution of a solute in a solvent increases the disorderliness (entropy) of the system because of the increase in the number of particles present. Hence once a solute in dissolved, the entropy of the system increases.

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