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nika2105 [10]
2 years ago
10

The vapor pressure of pure acetone (C3H6O) is 216 torr

Chemistry
1 answer:
svetoff [14.1K]2 years ago
6 0

The mole fraction of acetone (C₃H₆O) is 0.333

<h3>Data obtained from the question </h3>
  • Mole of C₃H₆O = 0.1 mole
  • Mole of CHCl₃ = 0.2 mole
  • Mole fraction of C₃H₆O =?

<h3>How to determine the mole fraction </h3>

Mole fraction of a substance can be obtained by using the following formula:

Mole fraction = mole / total mole

With the above formula, we can obtain the mole fraction of C₃H₆O as follow:

  • Mole of C₃H₆O = 0.1 mole
  • Mole of CHCl₃ = 0.2 mole
  • Total mole = 0.1 + 0.2 = 0.3 mole
  • Mole fraction of C₃H₆O =?

Mole fraction = mole / total mole

Mole fraction of C₃H₆O = 0.1 / 0.3

Mole fraction of C₃H₆O = 0.333

Learn more about mole fraction:

brainly.com/question/2769009

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A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the
alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.4 atm

P_2 = final pressure of gas = 1.0 atm

V_1 = initial volume of gas = 250 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 15^oC=273+15=288K

T_2 = final temperature of gas = 27^oC=273+27=300K

Now put all the given values in the above equation, we get:

\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}

V_2=625mL

Therefore, the volume of the bubble is, 625 mL

7 0
2 years ago
at a given tempature gas with a pressure of 150 kpa has volume of 0.8 l if the pressure decreases to 75kpa and the temperature r
vova2212 [387]
  <span>P*V/T=constant 
so P*V= constant*T 
if T doesn't change then 
P*V= constant 
so 150kPa*0.8L=75kPa*xL 
xL=150kPa*0.8L/75kPa=1.6L 
hope it help</span>
8 0
3 years ago
Given bh3, o3, co2, and nh3, which of these compound is an exception to the octet rules?.
Ghella [55]

Answer:

a. BH₃

Explanation:

According to the octet rules, atoms reach stability when are surrounded by eight electrons in their valence shell when they combine to form a chemical compound.

From the options, the only compound in which the central atom does not meet the octet rules is BH₃. The central atom is boron (B), which has 3 electrons in its valence shell. When B is combined with hydrogen (H), 3 electrons from the 3 atoms of H are added. The total amount of electrons is 6, fewer than 8 electrons needed to meet the rule.

hope this helps

5 0
2 years ago
What is the net ionic equation for the reaction that occurs when aqueous solutions of koh and zncl2 are mixed?
Mademuasel [1]

Answer:

Net ionic equation:

Zn²⁺(aq)  +   2OH⁻(aq)     →    Zn(OH)₂(s)

Explanation:

Chemical equation:

ZnCl₂ + KOH    →   KCl  + Zn(OH)₂

Balanced chemical equation:

ZnCl₂ + 2KOH    →   2KCl  +Zn(OH)₂

Ionic equation;

Zn²⁺(aq)  + 2Cl⁻(aq)  + 2K⁺(aq)  +  2OH⁻(aq)     →   2K⁺(aq)  + 2Cl⁻(aq)   +Zn(OH)₂(s)

Net ionic equation:

Zn²⁺(aq)  +   2OH⁻(aq)     →    Zn(OH)₂(s)

The K⁺  and Cl⁻  are spectator ions that's why these are not written in net ionic equation. The  Zn(OH)₂ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

5 0
3 years ago
3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g
hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g  of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid =  diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol

5 0
3 years ago
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