Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
Answer:
- Volume = <u>2.0 liter</u> of 1.5 M solution of KOH
Explanation:
<u>1) Data:</u>
a) Solution: KOH
b) M = 1.5 M
c) n = 3.0 mol
d) V = ?
<u>2) Formula:</u>
Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:
<u>3) Calculations:</u>
- Solve for n: M = n / V ⇒ V = n / M
- Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter
You must use 2 significant figures in your answer: <u>2.0 liter.</u>
The nuclear reaction occurring is known as alpha-decay, and during this process, an alpha particle is released from a heavy radioactive nucleus to form a lighter more stable nucleus. The alpha particle is equivalent to a helium nucleus, which means it contains 2 protons and two neutrons (net charge of +2)
The decay equation is:
Rn → Po + α
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
