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Anit [1.1K]
1 year ago
11

A 609 ml sample of naoh has a ph of 13.5292. if 594 ml of distilled water was added to the initial naoh solution, what would the

new ph of the solution be g
Chemistry
1 answer:
Delvig [45]1 year ago
8 0

The new pH of the solution after the addition of water would be 13.22.

<h3>Total volume of the solution</h3>

V(total) = 609 ml + 594 ml

V(total) = 1203 ml

<h3>Concentration of water </h3>

C = (594 ml)/(1203 ml)

C = 0.493

pOH = -log(0.493)

pOH = 0.31

New pH of the solution = 13.5292 - 0.31 = 13.22

Learn more about pH here: brainly.com/question/13557815

#SPJ1

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Answer:

n = 7.86 mol

Explanation:

This question can be solved using the ideal gas law of PV = nRT.

Temperature must be in K, so we will convert 22.5C to 295 K ( Kelvin = C + 273).

R is the ideal gas constant of 0.0821.

(2.24atm)(85.0L) = n(0.0821)(295K)

Isolate n to get:

n = (2.24atm)(85.0L)/(0.0821)(295K)

n = 7.86 mol

8 0
3 years ago
Yasmin's teacher asks her to make a supersaturated saline solution. Her teacher tells her that the solubility of the salt is 360
Aleks [24]

Answer:

She can add 380 g of salt to 1 L of hot water (75 °C) and stir until all the salt dissolves. Then, she can carefully cool the solution to room temperature.  

Explanation:

A supersaturated solution contains more salt than it can normally hold at a given temperature.

A saturated solution at 25 °C contains 360 g of salt per litre, and water at 70 °C can hold more salt.

Yasmin can dissolve 380 g of salt in 1 L of water at 70 °C. Then she can carefully cool the solution to 25 °C, and she will have a supersaturated solution.

B and D are wrong. The most salt that will dissolve at 25 °C is 360 g. She will have a saturated solution.

C is wrong. Only 356 g of salt will dissolve at 5 °C, so that's what Yasmin will have in her solution at 25 °C. She will have a dilute solution.

3 0
3 years ago
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Accelerate, decelerate, and changing directions.
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3 years ago
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Answer:

<h3>Which of the following increases with the increase in the temperature in case of a liquid?</h3><h2><em>Va</em><em>pour</em><em> </em><em> </em><em>Pressure</em><em> </em></h2>

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