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podryga [215]
3 years ago
9

The molar masses of three gases, A, B, and C, are 4 grams, 16 grams, and 32 grams respectively. Which sentence represents the ef

fusion rates of these gases?
Chemistry
2 answers:
Colt1911 [192]3 years ago
7 0

B for plato users C<B<A

timurjin [86]3 years ago
3 0
This is where something called Graham's Law applies. Grahams law says that the rate of effusion is inversely proportional to the square root of the molar mass of a gas. What that means is that the less the molar mass of a gas is, the less quickly the gas effuses (effusion being the rate at which a gas can travel through a small hole). A great way to think about this is that effusion is thought about as though the gas is traveling through a small hole, so smaller particles would be able to go through it with greater ease than would a large particle. I don't know what particular sentences the question asks for are, but the answer should be that gas A (molar mass 4) has the greatest effusion rate, gas B (molar mass 16) has the second fastest effusion rate, and gas C (molar mass 32) has the slowest effusion rate.
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For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and
avanturin [10]

Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

-0.3378= log [Fe2+] /[0.040]

Antilog(-0.3378) = [Fe2+] /[0.040]

0.459= [Fe2+] /[0.040]

[Fe2+] = 0.459 × 0.040

[Fe2+] = 0.01836 M

7 0
3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0
3 years ago
What happens to the rate of reaction when you decrease the temperature of a chemical reaction?
Neporo4naja [7]
Depending on what chemicals you use it slows the reaction
4 0
3 years ago
Read 2 more answers
What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
3 years ago
Write the formula for the conjugate base for the following acids. <br>Thank you ​
Llana [10]

Answer:

See the answer and explanation below , please.

Explanation:

A conjugate base is defined as that formed after an acid donates its proton.

For each article, a continuation of the conjugate bases (highlighted in bold), for dissociation in water:

a) HF + H20 --> F- + H30+      

b) H20+ H20 --> OH-  + H30+

C)H2PO3-  + H20--> HPO3 2-   + H30+

d) HSO4- + H20 -->  SO4 2- + H30+

E) HCL02 + H20 --> CLO02 -   + H30+

7 0
4 years ago
Read 2 more answers
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