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Valentin [98]
3 years ago
15

A hiker has packed a bag of chips as a snack. At the start of the hike the pressure was 1.5 atm and the temperature was 35*C. At

the end of the hike the temperature has dropped to a 0.5*C. What is the pressure at the end of the hike? What would you expect to happen to the bag of chips?
Chemistry
1 answer:
Mazyrski [523]3 years ago
5 0
According to <span>Gay-Lussac's Law the temperature and Pressure are directly proportional to each other if the amount and volume of given gas are kept constant.
Mathematically for initial and final states it is expressed as,

                                          P</span>₁ / T₁  =  P₂ / T₂     ----- (1)
Data Given;
                  P₁  =  1.5 atm

                  T₁  =  35 °C + 273  =  308 K

                  P₂  =  ?

                  T₂  =  0 °C + 273  =  273 K

Solving Eq. 1 for P₂,

                                   P₂  =  P₁ T₂ / T₁

Putting values,
                                   P₂  =  (1.5 atm × 273 K) ÷ 308 K

                                   P₂  =  1.32 atm
Result:
           As the temperature is decreased so the pressure also decreases from 1.5 atm to 1.32 atm. Therefore the bag will contract.
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olga2289 [7]

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $\frac{V1M1}{V2}

Plugin the values as,

M2 = $\frac{150 \times 0.5}{60}

       = 1.25 M

So the molarity of the stock solution is 1.25 M.

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Why are there wires in circuits?what is their job/function?
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3 years ago
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78.6 grams of O2 and 67.3 grams of F2 are placed in a container with a volume of 40.6 L. Find the total pressure if the gasses a
saul85 [17]

1) List the known and unknown quantities.

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

2) Find the pressure of O2.

<em>2.1- List the known and unknown quantities.</em>

<em>Sample: O2.</em>

Mass: 78.6 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

<em>2.2- Convert grams of O2 to moles of O2.</em>

The molar mass of O2 is 31.9988 g/mol.

mol\text{ }O_2=78.6\text{ }g*\frac{1\text{ }mol\text{ }O_2}{31.9988\text{ }g\text{ }O_2}=2.46\text{ }mol\text{ }O_2

<em>2.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>2.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{O_2}=\frac{(2.46\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)}{40.6\text{ }L}P_{O_2}=1.57\text{ }atm

<em>The pressure of O2 is 1.57 atm.</em>

3) Find the pressure of F2.

<em>3.1- List the known and unknown quantities.</em>

<em>Sample: F2.</em>

Mass: 67.3 g.

Volume: 40.6 L.

Temperature: 43.13 ºC = 316.28 K.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

3.2- <em>Convert grams of F2 to moles of F2.</em>

The mmolar mass of F2 is 37.9968 g/mol.

mol\text{ }F_2=67.3\text{ }g\text{ }F_2*\frac{1\text{ }mol\text{ }F_2}{37.9968\text{ }g\text{ }F_2}=1.77\text{ }mol\text{ }F_2

<em>3.3- Set the equation.</em>

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1)

PV=nRT

<em>3.4- Plug in the known quantities and solve for P.</em>

(P)(40.6\text{ }L)=(1.77\text{ }mol\text{ }F_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(316.28\text{ }K)

<em>.</em>

P_{F_2}=\frac{(1.77molF_2)(0.082057L*atm*K^{-1}*mol^{-1})(316.28K)}{40.6\text{ }L}P_{F_2}=1.13\text{ }atm

<em>The pressure of F2 is 1.13 atm.</em>

4) The total pressure.

Dalton's law - Partial pressure. This law states that the total pressure of a gas is equal to the sum of the individual partial pressures.

<em>4.1- Set the equation.</em>

P_T=P_A+P_B

4.2- Plug in the known quantities.

P_T=1.57\text{ }atm+1.13\text{ }atmP_T=2.7\text{ }atm

<em>The total pressure in the container is </em>2.7 atm<em>.</em>

5 0
1 year ago
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IgorC [24]

Answer:

A. Chemical Reaction

Explanation:

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miss Akunina [59]

Molecular mass of C₂H₄ is,

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M = 28 g/mol

Moles of C₂H₄ in 5.6 g of C₂H₄ :

n = 5.6/28 mol

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Now, 1 mol of C₂H₄ contains 2 moles of carbon.

So, number of moles of carbon are :

n = 0.4 mol

We know, 1 mol of any atom contains 6.022 × 10²³ atoms.

So, number of carbon atoms are :

N = 0.4 \times 6.022\times 10^{23} \\\\N = 2.409 \times 10^{23}

Hence, this is the required solution.

4 0
3 years ago
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