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algol [13]
3 years ago
13

A scientist performs an experiment in which they create an artificial cell with a selectively permeable membrane through which o

nly water can pass. They inject a 5 M solution of glucose into the cell and then place the cell into a beaker of water. After an hour, what effect do you expect to observe?
Chemistry
1 answer:
lozanna [386]3 years ago
8 0

Answer:

Water moves into the cell

Explanation:

As shown in the question above, the cell is high in glucose and placed in a glass filled with water. This cell has a semi permeable membrane that allows only water to pass through, as the concentration of water within the cell is low, the cell will attempt to strike a balance with the medium it is inserted into. For this reason, what is likely to happen is the passage of water from the most concentrated to the least concentrated medium, that is, the water will pass from the cup to the cell.

water moves into the cell through osmosis.during osmosis water moves from a region of low concentration of solute to a region of high concentration of solute.the glucose introduced into the cell makes it more concentrated.

In this case the cell is hypertonic and water would enter into the cell through the semi permeable membrane.this membrane allows water to pass through but not glucose.this movement of water into the cell causes the cell to become turgid.

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What is the boiling point of a solution formed by dissolving 0.75 mol of KCl in 1.00 kg of water?
iren [92.7K]

The boiling point of water generally increases as the amount of impurities (which a solute like KCl technically can be thought of) dissolved increases. This relation can be quantified using the equation,

\Delta T_b = i \times K_b \times m

where \Delta{T}_{b} is the <em>change </em>in the water's boiling point (normally taken to be 100 °C), i is the Van 't Hoff factor (the number of particles a single formula unit of the solute dissociates into in water), K_b is the boiling point elevation constant, and m is the molality (moles of solute/kilogram(s) of solvent) of the solution.

We are forming a solution by dissolving KCl in water. KCl is an electrolyte that, in water, will dissociate into K⁺ and Cl⁻ ions. So, for every formula unit, KCl, we obtain two particles. Thus, the Van 't Hoff factor, or i, will be 2.

The molality of the solution can be calculated by dividing the number of moles of KCl by the mass of water in kilograms. Since we have 1.00 kg of water, we would be dividing 0.75 mol KCl by 1, giving us a molality (<em>m</em>) of 0.75 <em>m.</em>

We aren't provided the boiling point elevation constant for water. Several authoritative sources give the value 0.512 °C/<em>m</em>, so we will adopt that as our K_b.

Note: <em>m = mol/kg </em>as used in this problem.

Plugging everything in,

\Delta T_b = i \times K_b \times m \\\Delta T_b = 2 \times 0.512 \text{ } \frac{^oC}{mol/kg} \times 0.75 \text{ } \frac{mol}{kg} \\\Delta T_b = 0.768 \text{ } \mathrm{ ^oC}

As you can see, our change in boiling point is positive (the boiling point is elevated), and it is also quite modest. Taking 100 °C to be the boiling point of pure water, the boiling point of our solution would be 100 ⁰C + 0.768 ⁰C, or 100.768 ⁰C.

If we are considering significant figures, then we must give our answer to two significant figures (since 0.75 has two sig figs). We can regard the boiling point of water (100 ⁰C) as a defined value. Since our final answer is a sum, the boiling point of our solution to two significant figures would be 100.77 ⁰C.

4 0
3 years ago
When 20.4g of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride
vagabundo [1.1K]

Answer: True

Explanation:

\mathrm{The\ balanced \ chemical \ equation\ is\ :}\\$2 \mathrm{Na}{(s)}+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(s)

<u>For Na</u>

\\$Given mass $=20.4 \mathrm{~g}$ \ \ Molecular mass $=23 \mathrm{~g}\ \ $Number of moles $=\frac{\text { Given mass }}{\text { Molecular mass }}$$$\begin{aligned}&=\frac{20.4}{23} \\&=0.887 \text { moles }\end{aligned}$$Stoichiometric ratio of $\mathrm{Na}$ is same as that of $\mathrm{NaCl}$ i.e., $1: 1$.$\therefore$ Moles of $\mathrm{NaCl}=0.887$ motes

<u>For NaCl</u>

$$\\\begin{aligned}\text { Molecular mass } &=23+35.5 \\&=58.5 \mathrm{~g}\end{aligned}$$

\begin{aligned}\\& $\therefore$ Mass of $\mathrm{NaCl}$ produced will be=\text { No.of moles } \times \text { Molecular mass } \\&=0.887 \times 58.5 \\&=51.9 \\&\simeq 52 \mathrm{~g} \quad\end{aligned}$$

Therefore, it is true that when 20.4g of sodium metal are mixed with chlorine gas, 52.0 g of sodium chloride is produced

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galina1969 [7]

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