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wariber [46]
2 years ago
14

What will occur if solution A containing 400 mosmol/L nonpenetrating solute is separated by a biological membrane from solution

B containing 600 mosmol/L nonpenetrating solute?
Chemistry
1 answer:
Natali [406]2 years ago
7 0

Answer:

The volume of solution B will increase

Explanation:

If a solute is non-penetrating it means it cannot pass through a biological membrane. In this case we have two non-penetrating solutions with different osmolarity separated by biological membrane. Solvent from solution with lower osmolarity will tend to pass the membrane in order to equalize the solute concentrations on the two sides of the membrane. This process is called osmosis and it is spontaneous. Solvent moves through the membrane from a less concentrated solution (400 mosmol/L) into a more concentrated one (600 mosmol/L). Because of that, the volume of solution B will increase.

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2 years ago
If 455-ml of 6.0 M HNO3 is used to make 2.5 L dilution, what is the molarity of the dilution
Anna11 [10]
Hi there

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5 0
2 years ago
A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture
Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

n=\frac{PV}{RT}

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

n=(\frac{1.45*2.50}{0.0821*293})

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

CH_4+2O_2\rightarrow CO_2+2H_2O

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = 8.60g(\frac{1mol}{44g})

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

X=\frac{0.258}{2}

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = \frac{moles of Methane}{total moles}

mole fraction of methane = \frac{0.129}{0.151}

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

6 0
2 years ago
Is the ccl3– molecule polar or nonpolar? the molecule is nonpolar if the net molecular dipole moment is zero. the molecule is po
zimovet [89]
In the bonding of elements to form compounds, intermolecular forces must be in place. These forces are due to the difference in electronegativity. The force is directed towards the more electronegative element because it attracts more electrons towards itself. In this case, Chlorine is more electronegative than Carbon.

You have to make the Lewis structure as shown in the figure. Since there are 4 valence electrons for Carbon and 7 for Cl, the total electrons would be 4 +3(7) = 25. A polar molecule is when there is an imbalance of forces creating a dipole moment. In this case, the opposite Cl branches cancel out leaving one for for the upper Cl. There is an imbalance, therefore CCl3⁻ is a polar molecule. The basis is: t<span>he molecule is nonpolar if the net molecular dipole moment is zero.</span> 

4 0
3 years ago
A scientist makes an acid solution by adding drops of acid to 1.2 l of water. the final volume of the acid solution is 1.202 l.
Nataliya [291]

Answer:- 40 drops and the percentage of acid in acid solution is 0.166%.

Solution:- Acid is added drops wise to 1.2 L of water to make acid solution. The final volume of the acid solution is 1.202 L. From here we could calculate the volume of the acid added to water by subtracting water volume from final acid solution volume as:

volume of acid added = 1.202 L - 1.2 L = 0.002 L

It's given that the volume of each drop is 0.05 mL. let's convert the volume of acid added from L to mL.

0.002L(\frac{1000mL}{1L})

= 2mL

Now let's calculate the number of drops of acid added to water as:

2mL(\frac{1drop}{0.05mL})

= 40 drops

0.002 L of acid are present in 1.202 L of solution. It asks to calculate about what percent of acid solution is acid means how many L of acid are present in 100 L of acid solution. It's done as:

percentage of acid in acid solution = (\frac{0.002}{1.202})100

= 0.166%

So, 40 drops of acid are required and there is 0.166% of acid in acid solution.


5 0
2 years ago
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