A weak acid is one which:

Rank the following substances/solutions in order of lowest boiling point to highest boiling point where 1 has the lowest boiling

poizon [28]

Answer:

1) pure water

2) 0.75 m CaCl2

3) 1.0 m NaCl

4) 0.5 m KBr

5) 1.5 m glucose (C6H12O6)

Explanation:

Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.

The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.

The order of decreasing hoping point elevation is;

1) 0.75 m CaCl2

2) 1.0 m NaCl

3) 0.5 m KBr

4) 1.5 m glucose (C6H12O6)

5 0

3 years ago

(6 pt) Match each word from Column A with the correct definition from Column B. Column A Column B 1. element 2. compound 3. reac

Marat540 [252]

1. D
2. C
3. F
4. E
5. A
6. B

3 0

3 years ago

2. Determine the molarity of the NaOH solution in each trial. a. Trial 1 Molarity: b. Trial 2 Molarity: 3. Calculate the average

butalik [34]

Answer:

This question is incomplete

Explanation:

This question is incomplete but...

1) You can calculate the molarity of the NaOH for each trial by following the steps below.

The formula for Molarity (M) is

M = number of moles (n) ÷ volume (V)

where the unit of volume must be in Litres or dm³

The unit of molarity is mol/dm³ or mol/L or molar conc (M)

The final answer must have the unit of molarity

If the number of moles is not provided, look out for the mass of NaOH used and then calculate your number of moles (n) as

n = mass of NaOH used ÷ molar mass of NaOH

Where the atomic mass of sodium (Na) is 23, oxygen (O) is 16 and hydrogen (H) is 1. Hence, molar mass for NaOH is 23 + 16 + 1 = 40 g/mol

n = mass of NaOH used ÷ 40

2) Average Molarity will be (Trial 1 Molarity +Trial 2 Molarity) ÷ 2

Answer must be in mol/dm³ or mol/L or M

3) Label the volumentric flask containing the NaOH solution with the answer gotten from (2) above

3 0

3 years ago

Can you help me please

den301095 [7]

Type in the Eye lins it will help

4 0

3 years ago

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t

Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0

3 years ago