Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = = 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
The average pea weighs between 0.1 and 0.36 grams.
If we take the lower value (0.1 g/pea), the number of peas in 454 g is:
If we take the higher value (0.36 g/pea), the number of peas in 454 g is:
In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.
You can learn more about conversion factors here: brainly.com/question/1844638
Answer:
0.01M = [H⁺]; 1x10⁻¹²M = [OH⁻]; Ratio is: 1x10¹⁰
Explanation:
pH is defined as -log [H⁺]
For a pH of 2 we can solve [H⁺] as follows:
pH = -log [H⁺]
2 = -log [H⁺]
10^-2 = [H⁺]
<h3>0.01M = [H⁺]</h3>
Using Keq of water:
Keq = 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 0.01M = [OH⁻]
<h3>1x10⁻¹²M = [OH⁻]</h3><h3 />
The ratio is:
[H⁺] / [OH⁻] = 0.01 / 1x10⁻¹² =
<h3>1x10¹⁰</h3>
The activation energy barrier is 40.1 kJ·mol⁻¹
Use the Arrhenius equation
Answer:
2
Explanation:
The number of carbon atoms that are sp²-hybridized in this alkene is 2
Because all the single bonded carbon atoms in the alkene are sp²-hybridized
There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital
attached below is the some part of the solution