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olga_2 [115]
4 years ago
8

A pressure cooker contains 0.355 mole of air, which occupies a volume of 5.68 liters. The temperature inside the pressure cooker

is 394 K. What is the absolute pressure of the air in the pressure cooker? The pressure of the air is_ pascals. Express you answer to three significant figures.
Chemistry
1 answer:
sergejj [24]4 years ago
8 0

Answer:

We will use ideal gas equation in this problem

PV = nRT

Where

P = pressure of gas or air = ?

V = volume of air = volume of pressure cooker = 5.68 L

T = temperature = 394 K

n = moles of air = 0.355

R = ideal gas equation = 0.0821 L atm / mol K

Putting values

Pressure = nRT / V = 0.355 X 0.0821 X 394 / 5.68 = 2.02 atm.

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What mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv
statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0
3 years ago
Nguyên tử nguyên tố sắt có 26e<br> a. Viết cấu hình e của sắt<br> b. Sắt thuộc nguyên tố s,p,d,f
vichka [17]

Answer:

1s2 2s2 2p6 3s2 3p6 3d6 4s2

Explanation:

3 0
3 years ago
Gases with high molecular weights diffuse more slowly than gases with lower molecular weights.
Rom4ik [11]
The answer to this item is TRUE. This can be explained through the Graham's law. This law states that the rate at which gases diffuse is inversely proportional to the square root of their densities which is also related to their molecular masses. 
7 0
3 years ago
Read 2 more answers
I need help with 4, 5, 8, 9, and 6. Quickly I need it before class starts. Worth points!!!!!! HelP
GalinKa [24]

Answer:

4. 264.6J

5. 37.5J

6. 96J

7. 55Watts

8. 77.14m

9. 6s

10. 750Watts

Explanation:

4). Mechanical energy (potential energy) = mass (m) × acceleration due to gravity (g) × height (h)

m = 3kg, h = 9m, g = 9.8m/s²

P.E = 3 × 9 × 9.8

= 264.6J

5). Kinetic energy (K.E) = 1/2 × m × v²

Where;

m = mass (kg) = 3kg

v = velocity (m/s) = 5m/s

K.E = 1/2 × 3 × 5²

K.E = 1/2 × 3 × 25

K.E = 1/2 × 75

K.E = 37.5J

6). Work done (J) = Force (N) × distance (m)

Force = 12N, distance = 8m

Work done = 12 × 8

= 96J

7). Power = work done (J) ÷ time (s)

Work done = 550J, time = 10s

Power = 550/10

= 55Watts.

8). Work done = force (F) × distance (m)

Work done = 540J, force = 7N, distance = ?

540 = 7 × d

540 = 7d

d = 540/7

d = 77.14m

9). Power = work done (J) ÷ time (s)

Work done = 300J, time = ?, Power = 50Watts.

50 = 300/t

50t = 300

t = 300/50

t = 6seconds.

10). Power = work done (J) ÷ time (s)

This means that;

Power = force × distance / time

Force = 300N, distance = 5m, time = 2s

Power = 300 × 5 ÷ 2

Power = 1500 ÷ 2

Power = 750Watts

3 0
3 years ago
Calculate the mass of water produced when 42 g of propane, c3h8, is burned with 115 g of oxygen
gavmur [86]
The balanced combustion reaction of propane, C₃H₈, is

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane

Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen

Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen

Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.

Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams
6 0
4 years ago
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