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3 years ago

Explain why a buffer can be prepared from a mixture of NH4CL and NaOH but not from NH3 and NaOH .

1 answer:

4 0

For a buffer, you need a weak acid and its conjugate base, or a weak base and its conjugate acid. When you mix NH4Cl (which is slightly acidic) with NaOH, you form NH4OH, which is basic. This gives you the acid/base mixture you need for a buffer.

NH3 and NaOH, on the other hand, are both bases, so you won't have the acid required to form a buffer system.

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Why is a catalyst required in the haber process?

kykrilka [37]

Answer:

The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency

The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of ammonia in the equilibrium mixture. Its only function is to speed up the reaction.

Explanation:

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4 years ago

How many moles of O₂ are needed to burn 2.5 moles of CH₃OH?

IrinaK [193]

Answer:

3.75 moles

Explanation:

The chemical equation is 2CH₃OH + 3O₂ -> 2CO₂ + 4H₂O

2 moles of CH₃OH are burned by 3 moles of O₂

For 2.5 moles of CH₃OH are burned by x moles of O₂

Let's solve for x :

2*x=2.5*3 => 2*x=7.5 => x=3.75 moles of O₂ are needed to burn 2.5 moles of CH₃OH

5 0

3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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3 years ago

Use Boyle's Law to explain what would happen to the volume of a helium- filled balloon if it was carried underwater by a diver t

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The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.

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3 years ago

All are true for the isomerase reaction of glucose-6-phosphate to fructose-6-phosphate except:

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B it is an aldose to ketose isomerization

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4 years ago

Explain why a buffer can be prepared from a mixture of NH4CL and NaOH but not from NH3 and NaOH .​

Explain why a buffer can be prepared from a mixture of NH4CL and NaOH but not from NH3 and NaOH .