70.33 L is the volume of 10 moles of a gas at 300 K held at a pressure of 3.5 atm.
<h3>What is volume?</h3>
Volume is the percentage of a liquid, solid, or gas's three-dimensional space that it occupies.
Liters, cubic metres, gallons, millilitres, teaspoons, and ounces are some of the more popular units used to express volume, though there are many others.
We will use ideal gas law to find the volume
PV = nRT
Can also be written as
V = (nRT)/P
Where,
P = pressure
V = volume
n = amount of substance
R = ideal gas constant
T = temperature
Here, we have given
P = 3.5 atm
V = to find
n = 10 moles
R = 0.08206 L⋅atm/K⋅mol
T = 300k
Lets substitute the values
V = (10 × 0.08206 × 300)/3.5
V = 70.33 L
Learn more about volume
brainly.com/question/463363
#SPJ10
Kelvin. Kelvin is the scientific term used in temperature senses, and was created so everybody around the world would have the same measurements and not get confused about whether it was Fahrenheit or Celsius.
Volume fraction = volume of the element / volume of the alloy
Volume = density * mass
Base: 100 grams of alloy
mass of tin = 15 grams
mass of lead = 85 grams
volume = mass / density
Volume of tin = 15g / 7.29 g/cm^3 = 2.06 cm^3
Volume of lead = 85 g / 11.27 g/cm^3 = 7.54 cm^3
Volume fraction of tin = 2.06 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.215
Volume fraction of lead = 7.54 cm^3 / (2.06 cm^3 + 7.54 cm^3) = 0.785
As you can verify the sum of the two volume fractions equals 1: 0.215 + 0.785 = 1.000
Answer:
Land resources such as gravel and bedrock are used to construct buildings, roads, and sidewalks. Land is also where city structures—such as buildings, roads, and sidewalks—are constructed. People use land resources such as bedrock and aggregate to construct buildings, roads, and sidewalks.
I hope this answer helps
Answer:
0.84 moles of oxygen are required.
Explanation:
Given data:
Mass of CO₂ produced = 37.15 g
Number of moles of oxygen = ?
Solution:
Chemical equation:
C + O₂ → CO₂
Number of moles of CO₂:
Number of moles = mass/molar mass
Number of moles = 37.15 g/ 44 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of oxygen and carbon dioxide.
CO₂ : O₂
1 : 1
0.84 : 0.84
0.84 moles of oxygen are required.
