It would be atomic masses of the same atoms, and that atoms will be isotopes.
When you assume that the gas is behaving ideally, the gas molecules are very far from each other that they do not have any intermolecular forces. If it behaves this way, you can assume the ideal gas equation:
PV = nRT, where
P is the pressure
V is the volume
n is the number of moles
R is a gas constant
T is the absolute temperature
When the process goes under constant pressure (and assuming same number of moles),
P/nR = T/V = constant, therefore,
T₁/V₁=T₂/V₂
If V₂ = V₁(1+0.8) = 1.8V₁, then,
T₂/T₁ = 1.8V₁/V₁
Cancelling V₁,
T₂/300=1.8
T₂ =540 K
If you do not assume ideal gas, you use the compressibility factor, z. The gas equation would now become
PV =znRT
However, we cannot solve this because we don't know the value of z₁ and z₂. There will be more unknowns than given so we won't be able to solve the problem. But definitely, the compressibility factor method is more accurate because it does not assume ideality.
Answer:
48 grams
Explanation:
The chemical equation for the reaction is the following:
2 H₂ + O₂ → 2 H₂O
That means that 2 moles of H₂ react with 1 mol of O₂ to produce 2 moles of H₂O. We convert the moles of oxygen (O₂) by using the molecular weight (MW) as follows:
MW(O₂) = 16 g/mol x 2 = 32 g/mol
mass of O₂ = 1 mol x 32 g/mol = 32 g
So, we have the following stoichiometric ratio: 32 g O₂/2 moles H₂. We have 3 moles of hydrogen (H₂), so we multiply the moles by the stoichiometric ratio to calculate how many grams are needed:
3 moles H₂ x 32 g O₂/2 moles H₂ = 48 g O₂
<em>Therefore, 48 grams of O₂ are needed to react with 3 moles of H₂.</em>
<h2>Increase of reaction rate</h2>
Explanation:
- It is observed that when the concentration of acetylcholine remains constant in the reaction of an aqueous solution, the speed of the enzyme-catalyzed reaction or the formation of the product increases with increasing concentrations of substrate.
- The reaction rate is directly proportional to the concentration of acetylcholine.
- At very low concentrations of acetylcholine, there is a small increase in the concentration of the substrate which results in a large increase of the rate in reaction.
Answer:
Because we need to dispense 4.7 mL, the volume reading in the pipet is the 5.3 mL line.
Explanation:
First we use C₁V₁=C₂V₂ in order to <u>calculate the required volume of concentrated HCl</u> (V₁):
12.85 M * V₁ = 0.600 M * 100 mL
V₁ = 4.7 mL
<u>So we need to dispense 4.7 mL of the concentrated HCl solution</u>. The mark in the pipet that would contain that volume would be 10.0 - 4.7 = 5.3 mL
