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eimsori [14]
2 years ago
8

The mole fraction of iodine, i2, dissolved in dichloromethane, ch2cl2, is 0.115. what is the molal concentration, m, of iodine i

n this solution?
Chemistry
1 answer:
german2 years ago
5 0
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:

0.115 mol I₂

1 - 0.115 = 0.885 mol CH₂Cl₂

We need moles of solute, which we have, and must convert our moles of solvent to kg:

0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂

We can now calculate the molality:

m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂

The molality of the iodine solution is 1.53.
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Convert 5.0x10^24 molecules to liters.
makvit [3.9K]
Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!
5 0
2 years ago
We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?
Anna11 [10]
M=D*V
D=620 g/cm³
V=75 cm³

m= 620 g/cm³ * 75 cm³=46500 g
m=46500g
8 0
2 years ago
An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:
aalyn [17]

Answer:

2,4,4-trimethyl-2-pentene yields mixture of (CH_{3})_{2}C=O and (CH_{3})_{3}CHO

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, Me_{2}S etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of (CH_{3})_{2}C=O and (CH_{3})_{3}CHO

Reaction steps are shown below.

8 0
2 years ago
A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr
zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0
2 years ago
At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?
Gnom [1K]
The temperature that  would  the volume of a gas  be 0.550l  if  it  had a volume of 0.432 L  at  -20.0  c is calculated  using the Charles law formula

that is   v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l

by  making T1  the subject of the formula  T1= V1T2/V2


T1=  (0.55lL x253)/  0.432 l = 322.11 K  or  322.11-273 = 49.11 C
8 0
2 years ago
Read 2 more answers
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