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3 years ago

During active transport, substances move from regions of blank concentration to regions of blank concentration

1 answer:

4 0

Active transport runs counter to facilitated diffusion. In active transport, molecules move against the concentration gradients, running from areas of lower concentration to areas of higher concentration. This is where energy is used.

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A solution has a volume of 2.0 L and contains 36.0 grams of glucose (C6H12O6). If the molar mass of glucose is 180 g/mol, what i

zysi [14]

36.0 g of glucose divided by 180 g/mol = 0.200 moles of glucose

find molarity
0.200 moles of glucose / 2 liters = 0.100 molar solution

(hope this helps)

6 0

3 years ago

Complete the sentences to best explain the ranking.

Jobisdone [24]

Answer:

dipole-dipole forces, ion-dipole forces, higher molar mass, hydrogen bonding, stronger intermolecular forces

Explanation:

1. H₂S and H₂Se exhibit the following intermolecular forces: dipole-dipole forces and ion-dipole forces. These molecules have a bent geometry, thus, a dipolar moment which makes them dipoles. When they are in the aqueous form they are weak electrolytes whose ions interact with the water dipoles

2. Therefore, when comparing H₂S and H₂Se the one with a higher molar mass has a higher boiling point. In this case, H₂Se has a higher boiling point than H₂S due to its higher molar mass.

3. The strongest intermolecular force exhibited by H₂O is hydrogen bonding. This is a specially strong dipole-dipole interaction in which the positive density charge on the hydrogens is attracted to the negative density charge on the oxygen.

4. Therefore, when comparing H₂Se and H₂O the one with stronger intermolecular forces has a higher boiling point. That's why the boiling point of H₂O is much higher than the boiling point of H₂Se.

4 0

3 years ago

Which element probably has the most properties in common with lead (Pb)?

Fed [463]

Answer:

Here is your answer: Tin,

Explanation:

Which of the following elements probably has the most properties in common with lead (Pb)? Tin (Sn).

8 0

3 years ago

A 2.350×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then b

sveta [45]

Answer:

Part A: m = 0.02356 mol/kg = 0.02356 m

Part B: Xsolute = 4.243×10⁻⁴

Part C: % m/m = 0.1376%

Part D: ppm = 1,376 ppm

Explanation:

1. Data:

a) M = 2.350×10⁻² M

b) V sol = 1.000 L

c) V H₂O = 994.4 mL = 0.9944 L

d) d H₂O = 0.9982 g/mL

2. Formulae

M = n solute / V sol (L)
m = n solute / Kg solvent
X solute = n solute / N total
% m/m = (mass of solute / mass of solution) × 100
ppm = (mass of solute / mass of solution) × 1,000,000
density = mass in grams / volume in mL

3. Solution

Part A: Calculate the molality of the salt solution.

m = n solute / Kg solvent

i) M = n solute / V sol (L) ⇒ n solute = M × V sol (L)

⇒ n solute = M = 2.350×10⁻² M × 1.000 L = M = 2.350×10⁻² mol

ii) density H₂O = mass H₂O / volume H₂O

⇒ mass H₂O = density H₂O × volume H₂O

⇒ mass H₂O = 0.9982 g/mL × 999.4 mL = 997.6 g

iii) kg H₂O = 997.6 g / (1,000 g/Kg) = 0.9976 kg

iv) m = 2.350×10⁻² mol / 0.9976 kg = 0.02356 mol/kg = 0.02356 m

Part B: Calculate the mole fraction of salt in this solution.

X solute = n solute / N total

i) n solute = 2.350×10⁻² mol

ii) n solvent = n H₂O = mass H₂O in grams/ molar mass H₂O

⇒ 997.6 g / 18.015 g/mol = 55.38 mol

iii) X solute = 2.350×10⁻² mol / 55.38 mol = 4.243×10⁻⁴

Part C: Calculate the concentration of the salt solution in percent by mass.

% m/m = (mass of solute / mass of solution) × 100

i) molar mass = mass in grams / molar mass

⇒ mass of solute = mass of NaCl = n solute × molar mass NaCl

⇒ mass of solute = 2.350×10⁻² mol × 58.44 g/mol = 1.373 g

ii) % m/m = (1.373 g / 997.6 g) × 100 = 0.1376%

Part D: Calculate the concentration of the salt solution in parts per million.

ppm = (mass of solute / mass of solution) × 1,000,000

i) ppm = ( (1.373 g / 997.6 g) × 1,000,000 = 1,376 ppm

5 0

3 years ago

If the pH of a weak acid solution is 2.500 and the solution has a concentration of 0.100M, what is the Ka of the weak acid HA?

Paul [167]

Answer:

The Kₐ of the weak acid is 1.033×10⁻⁴

Explanation:

The dissociation of a weak acid in aqueous solution is limited to about 5 to 10%

The acid dissociation reaction is given as follows;

HA (aq) + H₂O (l) → H₃O⁺(aq) + A⁻ (aq)

Given that the pH = 2.5, we have

pH = -log₁₀[H₃O⁺] = 2.5

∴ [H₃O⁺] = 10^(-2.5) = 0.0031623

$K_a = \dfrac{[H_3O^+][A^-]}{[HA]}$

Kₐ = [H₃O⁺][A⁻]/[HA] = (0.0031623^2)/(0.1 - 0.0031623) = 1.033×10⁻⁴

The acid dissociation constant, Kₐ for weak acid is very low as obtained

3 0

3 years ago