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SIZIF [17.4K]
3 years ago
8

A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure

effuses in 2.42 min. What is the molar mass of the unknown gas
Chemistry
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

<em>Molar mass of the gas is 0.0961 g/mol</em>

<em></em>

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of H_{2} effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

\frac{R_{g} }{R_{h} } = \sqrt{\frac{M_{h} }{M_{g} } }

where

R_{h} = rate of effusion of hydrogen gas

R_{g} = rate of effusion of unknown gas

M_{h} = molar mass of H2 gas

M_{g} = molar mass of unknown gas

substituting values, we have

\frac{11.1 }{2.42 } = \sqrt{\frac{2 }{M_{g} } }

4.587 = \sqrt{\frac{2 }{M_{g} } }

\sqrt{M_{g} } = \sqrt{2}/4.587

\sqrt{M_{g} } = 0.31

M_{g} = 0.31^{2} = <em>0.0961 g/mol</em>

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A certain half-reaction has a standard reduction potential +0.80 V . An engineer proposes using this half-reaction at the anode
Strike441 [17]

Answer:

a. Minimum 1.70 V

b. There is no maximum.

Explanation:

We can solve this question by remembering that the cell potential is given by the formula

ε⁰ cell = ε⁰ reduction -  ε⁰  oxidation

Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the  oxidized species  0.80 V, thus

ε⁰ reduction -  ε⁰  oxidation ≥  ε⁰ cell

Since ε⁰  oxidation is by definition the negative of ε⁰ reduction , we have

ε⁰ reduction - ( 0.80 V )  ≥  0.90 V

⇒ ε⁰ reduction  ≥ 1.70 V

Therefore,

(a) The minimum standard reduction potential is 1.70 V

(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V

8 0
3 years ago
Type the correct answer in the box. Express your answer to two significant figures. An industrial vat contains 650 grams of soli
nikklg [1K]

88.98 %

The Balance Chemical Equation is as follow,

                                  2 HCl + Pb(NO₃)₂     →    2 HNO₃ + PbCl₂

According to equation,

          331.2 g (1 mole) Pb(NO₃)₂ produces  =  278.1 g (1 mole) PbCl₂

So,

                 870 g of Pb(NO₃)₂ will produce  =  X g of PbCl₂

Solving for X,

                     X  =  (870 g × 278.1 g) ÷ 331.2 g

                     X  =  730.5 g of PbCl₂

Therefore,

               Theoretical Yield =  730.5 g

Also as given,

                Actual Yield =  650 g

So using following formula for percentage yield,

                        %age Yield  =  (Actual Yield / Theoretical Yield) × 100

Putting values,

                        %age Yield  =  (650 g / 730.5 g) × 100

                        %age Yield  =  88.98 %

Brianliest please and thank you.

6 0
4 years ago
Using the molecular orbital model to describe the bond- ing in F2????, F2, and F2????, predict the bond orders and the relative
masya89 [10]

Answer: F2 : bond order= 1.0

F2+: bond order = 1.5

F2- : bond order = 0.5

Explanation:

1. Starting with F2+

The configuration gives;

F2+ = 9F = 1S2.2S2.2P5

= 9F+ = 1S2.2S2.2P4 (this shows it gives out an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py1

The number of Electrons = (9*2) – 1 = 18 -1 = 17

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 7

Bond order = (10-7)/2 = 3/2 = 1.5

Number of unpaired electrons = 1

2. Starting with F2

The configuration gives;

F2 = 9F = 1S2.2S2.2P5

9F = 1S2.2S2.2P5 (this shows no loss of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2

The number of Electrons = (9*2) = 18 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 8

Bond order = (10-8)/2 = 2/2 = 1.0

Number of unpaired electrons = 0

3. Starting with F2-

The configuration gives;

F2- = 9F = 1S2.2S2.2P5

10F--= 1S2.2S2.2P6 (this shows an addition of an electron)

Electronic configuration = σ 1S2σ*1S2σ2S2σ*2S2σ2Pz2 π2Px2 = π2Py2 π*2Px2 = π*2Py2 σ*2Pz

The number of Electrons = (9*2) + 1 = 19 electrons

Bonding electrons (Nb) = 10

Antibonding electrons (Na) = 9

Bond order = (10-9)/2 = 1/2 = 0.5

Number of unpaired electrons = 1

To get the order of bond as well as length, we know that;

Bond order directly proportional to 1/ Bond length

Therefore the Ascending Bond length = F2+ ˂ F2 ˂ F2-

3 0
3 years ago
A sample of 23.2 g of nitrogen gas is reacted with
slavikrds [6]

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

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

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

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

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

5 0
3 years ago
True or False: Your blood is a neutral solution.
MaRussiya [10]

Answer:

True

Explanation:

Neutral solutions maintain a pH of 7. Water and human blood are great examples of neutral solutions. Acids mixed with bases can be neutralized and given a pH of 7.

8 0
3 years ago
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