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Svetlanka [38]
1 year ago
14

What is the ratio of effusion rates for the lightest gas, h2, to the heaviest known gas, uf6?

Chemistry
1 answer:
andriy [413]1 year ago
8 0

The ratio of effusion rates for the lightest gas H₂ to the heaviest known gas UF₆ is 13.21 to 1

<h3>What is effusion?</h3>

Effusion is a process by which a gas escapes from its container through a tiny hole into evacuated space.

Rate of effusion ∝ 1/√Ц, (where Ц is molar mass)

Rate H₂ = 1/√ЦH₂

Rate UF₆ = 1/√ЦUF₆

Therefore, Rate H₂/ Rate UF₆ = √ЦH₂/√ЦUF₆

ЦH₂= 2.016 g/mol

ЦUF₆= 352.04 g/mol

Rate H₂ / Rate UF₆ = √352.04/√2.016 = 18.76/1.42

Rate H₂ / Rate UF₆ = 13.21

Therefore, H₂ is lower mass than UF₆. Thus H₂ gas will effuse 13 times more faster than UF₆ because the most probable speed of H₂ molecule is higher; therefore, more molecules escapes per unit time.

learn more about effusion rate: brainly.com/question/28371955

#SPJ1

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What is an example of equilibrium?
frez [133]

A good example of equilibrium would be the mixing of oil and water in a closed container.

<h3>What is chemical equilibrium?</h3>

Chemical equilibrium is a condition in which the concentrations of components of a chemical reaction remain unchanged and have no tendency to change.

Of all the options, the only one where the concentrations of the component reactants cannot change is a mixture containing oil and water in a closed container.

Oil and water are immiscible and thus, their concentrations remain constant.

More on chemical equilibrium can be found here: brainly.com/question/4289021

#SPJ1

5 0
2 years ago
You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H
KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+    pKa1 = 2.14

H2PO4 = HPO4 + H+       pKa2 = 6.86

HPO4 = PO4 + H+      pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

Download odt
7 0
3 years ago
What is the pH of the solution if the [H+] is 1 x 10-12 ?
jarptica [38.1K]

Answer:

<h2>12</h2>

Explanation:

The pH of a solution can be found by using the formula

pH = - log [ {H}^{+} ]

From the question we have

pH =  -  log(1 \times  {10}^{ - 12} )  \\

We have the final answer as

<h3>12</h3>

Hope this helps you

5 0
3 years ago
PLEASE HELP URGET!!!!
strojnjashka [21]

Answer:

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Explanation:

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5 0
3 years ago
How many moles in 6.57 x 10^24 formula units of NaCl?
laiz [17]

Answer:

3.955*10^48

Explanation:

1 mole of a substance gives 6.02*10^23/6.57*10^24 will give x then cross multiply the answer. is 3.955*10^48

8 0
3 years ago
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