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fgiga [73]
3 years ago
6

Use the reaction:

Chemistry
2 answers:
damaskus [11]3 years ago
5 0
The answer is 34.2 mL
goblinko [34]3 years ago
3 0

Answer:

It’s either 34.2 OR 34.3 either would be correct!

Explanation:

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Potassium carbonate dissolves as follows:
lara [203]

The 0.25 volume in liters of 1.0 M K_{2}CO_{3} solution is required to provide 0.5 moles of K(aq).

Calculation,

The Potassium carbonate dissolves as follows:

K_{2}CO_{3}(s) → 2K(aq) +CO_{3}^{-2} (aq)

The mole ratio is 1: 2

It means, the 1 mole K_{2}CO_{3} required to form 2 mole of K(aq).

To provide 0.5 mole of K(aq) = 1 mole ×0.5 mole /2 mole required by K_{2}CO_{3}.

To provide 0.5 mole of K(aq) ,0.25 mole required by K_{2}CO_{3}.

The morality of  K_{2}CO_{3} = 1 M = number of moles / volume in lit

The morality of  K_{2}CO_{3} =   1 M = 0.25 mole/ volume in lit

Volume in lit = 0.25 mole / 1 M = 0.25 mole/mole/lit = 0.25 lit

learn about moles

brainly.com/question/26416088

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5 0
2 years ago
The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van
Goshia [24]

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: ΔT_{b}=i K_{b}.m, where ΔT_{b}= elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

K_{b}= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i=\frac{2.5}{0.512 X 1.83}

i=2.66= 2.7 (approx.)


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3 years ago
Which natural law do biogeochemical cycles address?
emmainna [20.7K]
Im not completely sure but It might be Cycling of Matter
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26. B
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Material use make car tyres and chewing gum?
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