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defon
3 years ago
6

A gas made up of homonuclear diatomic molecules escapes through a pinhole times as fast as gas. Write the chemical formula of th

e gas. "0.238"
Chemistry
1 answer:
Art [367]3 years ago
6 0

Answer:

The answer to the question is

Chlorine Cl₂

Explanation:

Graham's Law of effusion states that the rate of effusion of  a gas is inversely proportional to the square root of the molar mass

\frac{R_{1} }{R_{2}} = \sqrt{ \frac{M_{2}}{M_{1}} }

0.238 = \sqrt{ \frac{4}{M_{1}} }

Therefore M₁ = 4÷0.238² = 70.61

Hence the mass of the gas is 70.61, that is X₂ = 70.61  or the molar mass of the element X = 70.61÷2 = 35.308 ≅35.453 Hence the gas is chlorine Cl₂

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The larger the molecules of a substance, the the London forces between them. A larger molecule has more electrons and a greater
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Answer:

True

Explanation:

All the above statements buttress the fact that the larger molecule, the greater the magnitude of London forces between the molecules. Each of the statements above is a confirmation/explanation of this general rule.

6 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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The Dome of the Rock and Al-Aqsa Mosque are located in what city?
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They are located in Jerusalem.
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This is the energy that all moving objects have. 1.
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Every moving object has kinetic energy.
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Why doesn't gold react with oxygon
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Give me a thanks I been helped a lot of people and non of them is thanks me.

<span>Gold have a single outer electron. This seems disadvantageous, energy-wise, until you look at the orbitals the electrons are in. The lone electron is in an S-orbital. This orbital is thus half full (since s-orbitals can contain 2 electrons), whereas all the other inner orbitals in silver and gold are filled, and hence exceptionally stable. After a full orbital, the next most stable orbital is a half full one. </span>
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