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Oksi-84 [34.3K]
3 years ago
8

How many liters of 0.305 M K3PO4 solution are necessary to completely react with 187 mL of 0.0184 M NiCl2 according to the balan

ced chemical reaction?
Chemistry
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

6.55 mL of K₃PO₄ are required

Explanation:

We need to propose the reaction, in order to begin:

2K₃PO₄ (aq) + 3NiCl₂(aq) →  Ni₃(PO₄)₂ (s) ↓ + 6KCl (aq)

Molarity = mol/L (Moles of solute that are contained in 1 L of solution.)

M = mol / volume(L).  Let's find out the moles of chloride:

- We first convert the volume from mL to L → 187 mL . 1L / 1000mL = 0.187L

0.0184 M . 0.187L = 0.00344 moles of NiCl₂

Ratio is 3:2. Let's propose this rule of three:

3 moles of chloride react with 2 moles of phosphate

Then, 0.00344 moles of NiCl₂ will react with (0.00344 . 2) /3 = 0.00229 moles of K₃PO₄

M = mol / volume(L) → Volume(L) = mol/M

Volume(L) = 0.00229 mol / 0.350 M = 6.55×10⁻³L

We convert the volume from L to mL → 6.55×10⁻³L . 1000mL /1L = 6.55 mL

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The pKb of the base cyclohexamine, C6H11NH2, is 3.36. What is the pKa of the conjugate acid, C6H11NH3
bonufazy [111]

Answer:

10.64

Explanation:

Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.

C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻     pKb = 3.36

C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:

C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq)   pKa

We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.

pKa + pKb = 14.00

pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64

6 0
3 years ago
if the density of copper is 8.9g/cm^3 and the Volume of the sample is 23.4 mL what would be the mass? ​
Black_prince [1.1K]

Answer:

The mass of copper = 208.26 grams.

Explanation:

Density = mass / volume

8.9 = mass / 23.4

mass = 8.9 * 23.4

= 208.26 grams.

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A limiting factor helps an organism outcompete other organisms i think that’s the answer
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According to the law of conservation of mass, the total mass of the reactants is
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is equal to the total number of products

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