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sveta [45]
3 years ago
8

A cylinder with a moveable piston contains 92g of Nitrogen. The external pressure is constant at 1.00 atm. The initial temperatu

re is 200K. When the temperature is decreased by 85 K, by putting it in a lower temperature freezer, the volume will decrease, according to the Ideal Gas Law. Calculate the work for this process. Express your answer in J. The conversion factor between liter atmospheres and joules is 101.3 J
Chemistry
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:

Work done in this process = 4053 J

Explanation:

Mass of the gas = 0.092 kg

Pressure is constant = 1 atm = 101325 pa

Initial temperature T_{1} = 200 K

Final temperature T_{2} = 200 - 85 = 115 K

Gas constant for nitrogen = 297 \frac{J}{kg k}

When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.

⇒ V ∝ T

⇒ \frac{V_{2} }{V_{1} } = \frac{T_{2} }{T_{1} } ------------ ( 1 )

From ideal gas equation P_{1} V_{1} = m R T_{1} ------ (2)

⇒ 101325 × V_{1} = 0.092 × 297 × 200

⇒ V_{1} = 0.054 m^{3}

This is the volume at initial condition.

From equation 1

⇒ \frac{V_{2} }{0.054} = \frac{200}{115}

⇒ V_{2} = 0.094 m^{3}

This is the volume at final condition.

Thus the work done is given by W = P [V_{2} - V_{1} ]

⇒ W = 101325 × [ 0.094 - 0.054]

⇒ W = 4053 J

This is the work done in that process.

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The formula for a compound of Li₊ ions and Br_ ions is written LiBr. Why can't it be written Li₂Br? Why isn't it written BrLi?
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Answer:

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Explanation:

Lithium is in the first column of the periodic table, so it will have 1 valence electron.

Bromine is in the seventh column of the periodic table, so it will have seven valence electrons.

They must combine in a way to reach 8.

When combining elements to form compounds, the "crisscross method" is used. Above Li would be a charge of +1, and above Br would be a charge of -1.

Cross the 1 from the top of Li to the bottom of Br, and so there is 1 Br.

Cross the 1 from the top of Br to the bottom of Li, and so there is 1 Li.

It is not written BrLi because chemists decided to order them the other way. Technically speaking, it isn't wrong, but the positive charge is normally put on the left and the negative charge is normally put on the right.

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3 years ago
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What is the mass of 8.23 x 10^23 atoms of Ag
Gnom [1K]

Answer:

\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}

Explanation:

<u>Convert Atoms to Moles</u>

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}

8.23 *10^{23}  *\frac{1 \ mol \ Ag}{6.022*10^{23} }

Condense into one fraction.

\frac{8.23 *10^{23}  }{6.022*10^{23} } \ mol \ Ag

1.366655596 \ mol \ Ag

<u>Convert Moles to Grams</u>

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

  • Ag: 107.868 g/mol

Let's make another ratio using this information.

\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

Multiply by the number of moles we calculated.

1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}

The moles of silver cancel out.

1.366655596 *\frac {107.868 \ g \ Ag}{1 }

1.366655596 * {107.868 \ g \ Ag}

147.4184058 \ g\ Ag

<u>Round</u>

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

  • 147.<u>4</u>184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

147 \ g\ Ag

8.23 ×10²³ silver atoms are equal to approximately <u>147 grams.</u>

3 0
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Answer:

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Step 2: Calculate the number of moles of Mg

The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.

5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg

Step 3: Calculate the number of moles of Cl

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Step 4: Calculate the number of moles of O

The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.

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4 0
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