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Allushta [10]
2 years ago
6

What volume of 0.205 M K3PO4 solution is necessary to completely react with 134 mL of 0.0102 M NiCl2?

Chemistry
1 answer:
viktelen [127]2 years ago
5 0
 the balanced equation for the above reaction is as follows;
3NiCl₂ + 2K₃PO₄ ---> Ni₃(PO₄)₂ + 6KCl
stoichiometry of NiCl₂ to K₃PO₄ is 3:2
the number of NiCl₂ moles reacted - 0.0102 mol/L x 0.134 L = 0.00137 mol
according to molar ratio 
if 3 mol of NiCl₂ reacts with 2 mol of K₃PO₄
then 0.00137 mol of NiCl₂ reacts with - 2/3 x 0.00137 = 0.000911 mol of K₃PO₄

molarity of given K₃PO₄ solution - 0.205 M
there are 0.205 mol in 1000 mL 
therefore volume of 0.000911 mol - 0.000911 mol / 0.205 mol/L = 4.44 mL 
volume of K₃PO₄ needed is 4.44 mL
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7 0
1 year ago
A student is instructed to make 1 L of a 2.0 M solution of CaCl2 using dry salt. How should he do this?
Pachacha [2.7K]
<span>The student should follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>

</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L

Molarity = number of moles / volume of the solution

Hence, number of moles in 1 L = 2 mol

2. Find out the mass of dry CaCl</span>₂ in 2 moles.<span>

moles = mass / molar mass

Moles of CaCl₂ = 2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol

Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
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3. Weigh the mass accurately 

4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and finally wash the funnel and watch glass with de-ionized water. That water also should be added into the volumetric flask.

5. Then add some de-ionized water into the volumetric flask and swirl well until all salt are dissolved.

<span>6. Then top up to mark of the volumetric flask carefully. 
</span>
</span>
7. As the final step prepared solution should be labelled.
4 0
2 years ago
27. Sample of neon gas has a pressure of 4.1 atm and argon gas with a pressure of 3.2 atm are added to a
MrRa [10]

Answer:

Option C. 13.5 atm

Explanation:

From the question given above, the following data were obtained:

Pressure of Neon (Pₙₑ) = 4.1 atm

Pressure of Argon (Pₐᵣ) = 3.2 atm

Pressure of nitrogen (Pₙ₂) = 6.2 atm

Total pressure (Pₜ) =?

The total pressure in the container can be obtained by adding the pressure of the individual gases. This is illustrated below:

Pₜ = Pₙₑ + Pₐᵣ + Pₙ₂

Pₜ = 4.1 + 3.2 + 6.2

Pₜ = 13.5 atm

Therefore, the total pressure in the container is 13.5 atm

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3 years ago
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iogann1982 [59]

Answer:

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Explanation:

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