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tatyana61 [14]
3 years ago
12

Question 2

Chemistry
1 answer:
guajiro [1.7K]3 years ago
6 0

These questions all involve special cases of the ideal gas law, namely Boyle's, Charles', and Gay-Lussac's Laws. The ideal gas law relates together the absolute pressure (P), volume (V), the absolute temperature (T), and number of moles (n) of a gas by the following:

PV = nRT

where R is the universal gas constant.

The special cases of the ideal gas law are obtained by holding constant all but two of the variables of a gas.

Boyle's Law relates the pressure and volume of a given mass of gas at a constant temperature: PV = k or P₁V₁ = P₂V₂.

Charles' Law relates the volume and temperature of a given mass of gas at a constant pressure: V/T = k or V₁/T₁ = V₂/T₂.

Gay-Lussac's Law relates the pressure and temperature of a given mass of gas at a constant volume: P/T = k or P₁/T₁ = P₂/T₂.

Depending on what we're given and instructed to find in each question, we can figure out which law to use.

---

Question 2:

We are given the volume of a gas at some pressure, and we're to find the new volume of the gas at a different pressure. Here, we use Boyle's Law: P₁V₁ = P₂V₂ where P₁ = 60 atm, V₁ = 20.0 L, and P₂ = 30 atm. We want to find V₂, which we can determine by rearranging the equation into the form V₂ = P₁V₁/P₂. Note that pressure and volume are inversely related according to Boyle's Law; since we're decreasing the pressure, the new volume of the gas should be greater than the initial volume of 20.0 L.

V₂ = (60 atm)(20.0 L)/(30.0 atm) = 40.0 L.

So, at 30 atm, the balloon will have a volume of 40.0 L.

---

Question 3:

This is another Boyle's Law question. The standard pressure (our initial pressure) is 1 atm. Here, we are decreasing the volume of the gas, and we want to find the new pressure; the pressure of the gas should thus increase proportionally (the pressure will be greater than 1 atm). Rearranging Boyle's Law to solve for P₂, we get P₂ = P₁V₁/V₂.

P₂ = (1 atm)(8.00 L)/(3 L) = 2.67 atm.

So, the new pressure of the gas is 2.67 atm (or 3 atm if we're considering V₂ to comprise one significant figure).

---

Question 4:

Here, we are increasing the temperature of a gas at a known pressure, and we want to determine what the new pressure will be. This is a Gay-Lussac's Law question; from the law, we see that pressure and temperature are directly proportional. Since we're increasing the temperature of the gas, we should expect the pressure of the gas to be greater than the initial 200 atm. Gay-Lussac's Law rearranged to solve for P₂ gives us P₂ = P₁T₂/T₁. When working with gas laws, temperatures must be in Kelvin (°C + 273.15 = K). So, T₁ = 300.15 K, T₂ = 350.15 K, and P₁ = 200 atm.

P₂ = (200 atm)(350.15 K)/(300.15 K) = 233 atm.

So, if the temperature is increased from 27 to 77 °C, the pressure of the gas in the tennis ball will be 233 atm. Here, it's ambiguous how many sig figs to use; if we use one sig fig per P₁, then our P₂ would equal P₁, which I think would be an absurd for a question to ask for. I would stick with either 233 atm or 230 atm (following the two sig figs of the temperatures), or you may go with however you've been instructed.

---

Question 5:

This is a Charles' Law question; we're looking for the new volume of a gas when the temperature of the gas is increased. As was the case in Gay-Lussac's Law, the two parameters in Charles' Law—volume and temperature—are directly proportional. Since the temperature of the gas is increased, we should expect the new volume of the gas to also increase (V₂ will be greater than 5.00 L). Temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (5.00 L)(300.15 K)/(250.15 K) = 5.99 L.

---

Question 6:

Another Charles' Law question. As with question 5, we want to find the new volume of the gas after a change in temperature. This time, the final temperature is lower than the initial temperature, so we should expect that V₂ will be less than the initial 0.5 L. Again, temperatures in Kelvin.

V₂ = V₁T₂/T₁ = (0.5 L)(313.15 K)/(493.15 K) = 0.317 L.

So, the volume of the balloon when it is fully cooled by your refrigerator will be 0.317 L.

---

Question 7:

This is yet another Charles' Law question, and, again, we are solving for V₂ after a change in temperature. Since the final temperature is greater than the initial temperature, V₂ should be greater than 2.2 L. Again, the temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (2.2 L)(653.15 K)/(453.15 K) = 3.17 L.

The new volume of the gas is 3.17 L ≈ 3.2 L (two sig figs).

---

Question 8:

We return to Gay-Lussac's Law here; pressure and temperature are directly proportional, and the temperature of the gas is increased. Thus, P₂ should be greater than 3 atm. Again, remember that temperatures must be in Kelvin.

P₂ = P₁T₂/T₁ = (3 atm)(298.15 K)/(288.15 K) = 3.1 atm.

So, the pressure inside the can after the temperature rise is 3.1 atm. Not a big increase, but an increase nonetheless.

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?​
lesya692 [45]

Answer:

3.72 mol Hg

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Density = Mass over Volume

Explanation:

<u>Step 1: Define</u>

D = 13.6 g/mL

54.8 mL Hg

<u>Step 2: Identify Conversions</u>

Molar Mass of Hg - 200.59 g/mol

<u>Step 3: Find</u>

13.6 g/mL = x g / 54.8 mL

x = 745.28 g Hg

<u>Step 4: Convert</u>

<u />745.28 \ g \ Hg(\frac{1 \ mol \ Hg}{200.59 \ g \ Hg} ) = 3.71544 mol Hg

<u>Step 5: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.71544 mol Hg ≈ 3.72 mol Hg

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Answer:

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Explanation:

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Answer:

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Cho 6,72 gam Fe tan hết trong dung dịch H2SO4 đặc nóng dư thu được V lít khí SO2. Giá trị của V là
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