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max2010maxim [7]
3 years ago
14

How many grams of NaCl should be obtained to make 150 mL of 4.5 M solution

Chemistry
1 answer:
nadya68 [22]3 years ago
3 0

Answer:

39,5 grams should be obtained of NaCl

Explanation:

We calculate the weight of 1 mol of NaCl:

Weight 1 mol NaCl= Weight Na + Weight Cl= 23g + 35,5g=58,5 g/mol

4,5M--> 4,5 moles NaCl in 1000ml (1L) of solution

1000ml-----4,5 moles NaCl

150 ml------x=(150 mlx4,5 moles NaCl)/1000ml=0,675 moles NaCl

1 mol NaCl--------------58,5 grams

0,675molesNaCl---x= (0,675molesNaClx58,5 grams)/1 mol NaCl

x= 39,4879 grams

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Read 2 more answers
Calculate the number of moles in 2.5 particle NaOH<br>​
andreev551 [17]

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1.5055×10²⁴

Explanation:

................................

7 0
3 years ago
The K b Kb for an amine is 4.004 × 10 − 5 . 4.004×10−5. What percentage of the amine is protonated if the pH of a solution of th
Alexxandr [17]

Answer:

The percentage of amine is protonated is 63,07%

Explanation:

The reaction of an amine RNH₃ (weak base) with water is:

RNH₃ + H₂O ⇄ RNH₄⁺ + OH⁻

The kb is defined as:

kb = \frac{[RNH_{4}^+][OH^-]}{[NH_{3}]}

As kb = 4,004x10⁻⁵ and [OH⁻] is 10^{-(14-9,370)}=2,344x10^{-5}:

4,004x10^{-5} = \frac{[RNH_{4}^+][2,34x10^{-5}]}{[NH_{3}]}

1,708 = [RNH₄⁺] / [RNH₃] <em>(1)</em>

As the total amine is a 100%:

[RNH₄⁺] + [RNH₃] = 100% <em>(2)</em>

Replacing (1) in (2):

1,708 [RNH₃]+ [RNH₃] = 100%

2,708 [RNH₃] = 100%

[RNH₃] = 36,93%

Thus,

<em> [RNH₄⁺] = 63,07%</em>

The percentage of amine protonated is 63,07%

I hope it helps!

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4 years ago
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