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meaning:An acid is a chemical substance, usually a liquid, which contains hydrogen and can react with other substances to form salts. Some acids burn or dissolve other substances that they come into contact with.
Answer:
P₂ = 2.7 atm
Explanation:
Given data:
Initial temperature = 30°C
Initial pressure = 3.00 atm
Final temperature = -5°C
Final pressure = ?
Solution:
Initial temperature = 30°C = 30 + 273 = 303 K
Final temperature = -5°C = -5 + 273 = 268 K
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
3.0 atm / 303 K = P₂/268 K
P₂ = 3.0 atm × 268 K / 303 K
P₂ = 804 atm. K /293 K
P₂ = 2.7 atm
The concentration of the solution in mass percent is 60%.
<h3>What is the mass percent?</h3>
The concentration of a substance can be expressed in mass percent. This refers to the percentage of the solute that is contained in the solution.
Thus we can write;
Mass percent = Mass of solute/ Mass of solution * 100/1
Mass percent = 1.5kg/1.5kg + 1 Kg * 100/1
Mass percent = 60%
Hence, the concentration of the solution in mass percent is 60%.
Learn more about mass percent:brainly.com/question/5394922
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Answer:1.
1.Balanced equation
C4H10 + 9 02 ==> 5H20 +4CO2
2. Volume of CO2 =596L
Explanation:
1.Combustion of alkane is the reaction of alkanes with Oxygen. And the general equation for the combustion is;
CxHy +( x+y/4) O2 ==> y/2 02 + xCO2
Where x and y are number of carbon and hydrogen atoms respectively.
For butane (C4H10)
x=4 and y=10
Therefore
C4H10 + 9 02 ==> 5H20 +4CO2
2. Mass of butane = 0.360kg
Molar mass of C4H10 = ( 12×4 + 1×10)
= 48 +10=58g/mol= 0.058kg/mol
Mole = mass/molar mass
Mole = 0.360/0.058= 6.2moles
From the stoichiometric equation
1mole of C4H10 will gives 4moles of CO2
Therefore
6.2moles of C4H10 will gives 4 moles of 24.8 moles of CO2
Using the ideal gas equation
PV=nRT
P= 1.0atm
V=?
n= 24.8mol.
R=0.08206atmL/molK
T=20+273=293
V= 24.8 × 0.08206 × 293
V= 596L
Therefore the volume of CO2 produced is 596L
