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joja [24]
3 years ago
14

Imagine a beaker divided down the center by a rigid membrane that is freely permeable to water but impermeable to glucose. Side

1 contains a 10% glucose solution & side 2 contains the same volume of pure water. At equilibrium, what will be the situation?
1. water will continue to move from side 2 to side 1
2. the volume of liquid remain equal on both sides
3. water will continue to move from side 1 to side 2
4. the volume of liquid will be greater in side 2
5. the volume of liquid will be greater in side 1
Chemistry
1 answer:
damaskus [11]3 years ago
6 0

Answer:

D. The volume of liquid will be greater in side 1

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g A and B are both non-volatile molecular compounds. Substance A has a greater molar mass than substance B. If 50 g of substance
Doss [256]

Answer:

The correct answer is - option D.  the boiling point of solution A will be lower than the boiling point of solution B

Explanation:

Colligative properties such as a decrease in the freezing point of the solution, increase in the boiling point of substance, decrease in Lowering of vapor pressure, and other properties depend upon the number of molecules only.

In the given solution the equal amount of two solutions are mixed that is 50 grams however due to the difference in the molecular mass so the atoms present in both solution A and B will be different. It is known that the number of atoms of a substance is inversely proportional to the molecular mass of the particular substance.

As it is given that Solution B has a low molecular mass which means it has a high number of atoms that means its boiling point will be higher than solution A.

8 0
3 years ago
The number of grams per mole of something is also known as_______
Sergeeva-Olga [200]

Answer:

Molar mass

Explanation:

This is a counting unit which represents the mass in grams of a substance that make up one mole of the substance. This mass is calculated as follows:

Molar mass = Mass/ Number of moles

Units: grams/mol

5 0
3 years ago
Read 2 more answers
Help fast !
jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0
3 years ago
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe
Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

V_s=V_o-V_i=V_i\\V_o=2*V_i\\

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

r_o=r_i+e

The first equation becomes

\frac{4 \pi}{3}*r_o^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\\frac{4 \pi}{3}*(r_i+e)^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\(r_i+e)^{3}  = 2 *r_i^{3}  \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0
3 years ago
Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 %
Mademuasel [1]

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol

Step 2: Calculate the molar ratio

C = \frac{0.966}{0.643} =1.50\\\\P = \frac{0.643}{0.643} = 1.00\\\\O = \frac{2.58}{0.643} =4.00

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

7 0
3 years ago
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