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3 years ago

What does the constant bombardment of gas molecules against the inside walls of a container produce?

1 answer:

3 0

Pressure since pressure is defined as force per unit area and the molecules exert a force on the walls of the container when they bombard it

You might be interested in

A property of a substance that is the same no matter what size of sample is tested is _____.

KengaRu [80]

Answer:

B.) an intensive property

Explanation:

Bc it is a Characteristic Property that won't change no matter what size of sample is tested

6 0

3 years ago

How many L of carbon dioxide at 1.00 atm and 298.15 K are released from a car's engine upon consumption of a 60.0 L LIQUID tank

torisob [31]

Answer:

79,000 liters

Explanation:

1. Number of moles of gasoline

a) Convert 60.0 liters to grams

density = 0.77kg/liter

density = mass / volume

mass = density × volume

mass = 0.77kg/liter × 60.0 liter = 46.2 kg

46.2kg × 1,000g/kg = 46,200g

b) Convert 46,200 grams to moles

molar mass of C₈H₁₈ = 114.2 g/mol

number of moles = mass in grams / molar mass

number of moles = 46,200g / (114.2 gmol) = 404.55 mol

2. Number of moles of carbon dioxide, CO₂ produced

a) Balanced chemical equation (given):

C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)

b) mole ratio:

1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / x

Solve for x:

x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂

3. Convert the number of moles of carbon dioxide to volume

Use the ideal gas equation:

pV = nRT

V = nRT/p

p = 1 atm

T = 298.15K

n = 3,236.4 mol

R = 0.08206 (mol . liter)/ (K . mol)

Substitute and compute:

V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atm

V = 79,183 liter

Round to two significant figures (because the density has two significant figures): 79,000 liters ← answer

3 0

3 years ago

In an AP, sum of four consecutive terms is 28 and their sum of their squares is 276. Find the four numbers.

Aleks [24]

Four Consecutive terms are written as a-3, a-1, a+1,a+3. So

Sum of Terms : 4a=28
a=7

So the terms are 4,6,8,10

Verification : 4^2+6^2+8^2+10^2
= 216

4 0

3 years ago

A 5 g sample of lead (specific heat 0.129 /g˚C) is heated, then put in a calorimeter with 50 mL of water (specific heat 4.184 J/

Svetach [21]

Answer:

670.68°C

Explanation:

Given that:

volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g

specific heat (C) = 4.184 J/g˚C

Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C

The quantity of heat (Q) used to raise the temperature of a body is given by the equation:

Q = mCΔT

Substituting values:

Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J

Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.

-Q = mCΔT

-418.4 J = 5 g × 0.129 J/g˚C × ΔT

ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C

temperature change ΔT = final temperature - initial temperature

- 648 .68°C = 22°C - Initial Temperature

Initial Temperature = 22 + 648.68 = 670.68°C

4 0

3 years ago

what is the molarity of a concentrated hbr solution if the solution is 27.7% hbr by mass and has a density of 1.24 g/ml

givi [52]

Answer:
M = No.of moles of solute / Volume of solution in litres
HBr: H = 1 x 1 = 1 Br = 1 x 80
Molar mass : 81 g / mol
a. n = 27.7% / 100 =0.277 g / (81 g / mol) = 0.0034 mol

density = 1.24 g/ml ; D= m / v

b. 65.32 ml x 1 L / 1000 mL = 0.065 L


c. M= n / V
= 0.0034 mol / 0.065 L
M= 0.052 mol / L

The molarity of the concentrated HBr solution is 0.052 mol/L

3 0

4 years ago