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Advocard [28]
3 years ago
6

The image on the left shows a normal red blood cell, and the image on the right shows a cell that has been put into a new soluti

on. The circles represent the salt in the solutions.
Which statement describes the motion of the water molecules in this situation?

The water molecules move by active transport into the cell from low water concentration to high water concentration.
The water molecules move by osmosis into the cell from low water concentration to high water concentration.
The water molecules move by osmosis into the cell from high water concentration to low water concentration.
The water molecules move by active transport into the cell from high water concentration to low water concentration.

Chemistry
2 answers:
Neporo4naja [7]3 years ago
6 0

Answer:

I think its "The water molecules move by active transport into the cell from high water concentration to low water concentration."

vichka [17]3 years ago
5 0

Answer:

It's C

Explanation:

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3 0
3 years ago
Enter an equation showing how this buffer neutralizes added aqueous acid (HI). Express your answer as a chemical equation. Ident
andrey2020 [161]

Answer:

H^++NH_3\rightleftharpoons NH_4^+

Explanation:

Hello there!

In this case, since the buffer is not given, we assume it is based off ammonia, it means the ammonia-ammonium buffer, whereas the ammonia is the weak base and the ammonium ion stands for the conjugate acid. In such a way, when adding HI to the solution, the base of the buffer, NH3, reacts with the former to promote the following chemical reaction:

H^++NH_3\rightleftharpoons NH_4^+

Because the HI is totally ionized in solution so the iodide ion becomes an spectator one.

Best regards!

5 0
3 years ago
How many grams in 4.53 moles of MgBr2
musickatia [10]

Answer:

520.12554

Explanation:

8 0
2 years ago
Please help with #2 and #3
Vaselesa [24]

Answer:

2. V_2=17L

3. V=82.9L

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

\frac{V_2}{T_2}=\frac{V_1}{T_1}

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

V_2=\frac{V_1T_2}{T_1}

Therefore, we plug in the given data to obtain:

V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

V=\frac{nRT}{P}

Therefore, we plug in the data to obtain:

V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L

Best regards!

6 0
3 years ago
A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of
pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

<em>pH = 10.05</em>

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

<em />

I hope it helps!

3 0
3 years ago
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