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ahrayia [7]
3 years ago
14

1.2 g of sucrose is added to water with the final volume of 75 ml. What is the concentration of sucrose in %w/v?

Chemistry
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

16 g/L

Explanation:

Given that:-

Mass of sucrose added = 1.2 g

Volume of water = 75 mL = 0.075 L ( 1mL = 0.001 L )

%w/v is defined as the mass of the solution in 1 L of the solution.

So,

\%\ w/v=\frac{Mass_{solute}}{Volume_{Solution}}

\%\ w/v=\frac{1.2\ g}{0.075\ L}=16\ g/L

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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0
3 years ago
If 10.0 g water at 0.0°C is mixed with 20.0 g of water at 30.0°C what is the final temperature of the mixture?
Bond [772]

Answer:

1.67 gradius Celcius

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

7 0
2 years ago
The average kinetic energy of gas particles depends on which factor?
saveliy_v [14]
Only the temperature of gas
4 0
3 years ago
What is the definition of percent composition in your own words?
harkovskaia [24]

Answer:

percentage by mass of each element in a compound.

Explanation:

4 0
2 years ago
For each of the following sublevels, give the n and l values and the number of orbitals: (a) 5s; (b) 3p; (c) 4f
OverLord2011 [107]

Answer:

(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1

(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3

(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Sublevel number, 0 ≤ l ≤ n − 1

So,

(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1

(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3

(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals =  2l +1 = 7

6 0
3 years ago
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