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zhannawk [14.2K]
3 years ago
5

10. The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 63 s for the gas to effus

e, whereas nitrogen gas (N2) required 48 s. The molar mass of the unknown gas is closest to- A) 24 g/mol B) 37 g/mol C) 16 g/mol D) 32 g/mol E) 48 g/mol
Chemistry
1 answer:
ki77a [65]3 years ago
4 0

Answer:

chu papi

Explanation:

You might be interested in
Rank these acids according to their expected pKa values.ClCH2COOHClCH2CH2COOHCH3CH2COOHCl2CHCOOHIn order of highest pka to lowes
Ilia_Sergeevich [38]

Answer:

CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH  > ClCH₂COOH

Explanation:

Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.

Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.

  • The closer the substituent is to the carboxyl group, the greater is its effect.
  • The more substituents, the greater the effect.  
  • The effect tails off rapidly and is almost zero after about three C-C bonds.

CH₃CH₂-CH₂COOH —  EDG —                         weakest —  pKₐ = 4.82

      CH₃-CH₂COOH — reference —                                     pKₐ = 4.75

  ClCH₂-CH₂COOH — EWG on β-carbon— stronger —     pKₐ = 4.00

           ClCH₂COOH — EWG on α-carbon — strongest —  pKₐ = 2.87

6 0
3 years ago
Read 2 more answers
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
Does the motion of the moon affects how we see it from Earth? *
irina [24]
Yes that’s why we see it in different shapes all the time
7 0
2 years ago
Calculate the percent dissociation of benzoic acid C6H5CO2H in a 2.4mM aqueous solution of the stuff. You may find some useful d
Musya8 [376]

Answer:

34%

Explanation:

5 0
3 years ago
Identify one factor represented in the diagram that is used to classify the ten types of clouds
Ede4ka [16]
Low clouds
Stratus clouds are uniform grayish clouds that often cover the sky. Usually no precipitation falls from stratus clouds, but they may drizzle. When a thick fog “lifts,” the resulting clouds are low stratus. Nimbostratus clouds form a dark gray, “wet” looking cloudy layer associated with continuously falling rain or snow. They often produce light to moderate precipitation.

Middle clouds
Clouds with the prefix “alto” are middle-level clouds that have bases at 6,500 to 23,000 feet up. Altocumulus clouds are made of water droplets and appear as gray, puffy masses, sometimes rolled out in parallel waves or bands. These clouds on a warm, humid summer morning often mean thunderstorms by late afternoon. Altostratus clouds, gray or blue-gray, are made up of ice crystals and water droplets. They usually cover the sky. In thinner areas of them, the sun may be dimly visible as a round disk. Altostratus clouds often form ahead of storms that produce continuous precipitation.

High clouds
Cirrus clouds are thin, wispy clouds blown by high winds into long streamers. They are considered “high clouds,” forming at more than 20,000 feet. They usually move across the sky from west to east and generally mean fair to pleasant weather. Cirrostratus, thin, sheetlike clouds that often cover the sky, are so thin the sun and moon can be seen through them. Cirrocumulus clouds appear as small, rounded white puffs. Small ripples in the cirrocumulus sometimes resemble the scales of a fish, creating what is sometimes called a “mackerel sky.”

Vertical clouds
Cumulus clouds are puffy and can look like floating cotton. The base of each is often flat and may be only 330 feet above ground. The top has rounded towers. When the top resembles a cauliflower head, it is called “cumulus congestus.” These grow upward and if they continue to grow vertically can develop into a giant cumulonimbus, a thunderstorm cloud, with dark bases no more than 1,000 feet above ground and extending to more than 39,000 feet. Tremendous energy is released by condensation of water vapor in a cumulonimbus. Lightning, thunder and violent tornadoes are associated with them.
6 0
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