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Maksim231197 [3]
1 year ago
15

A solute that has been dissolved in a solvent ________.

Chemistry
1 answer:
grigory [225]1 year ago
8 0

Answer:

a solution: for example when sugar is dissolved in water it becomes a sugar solution

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What's the while compound called knowing all those lines are carbon​
Kisachek [45]

Answer:

<u>2-chlorohexane</u>

Explanation:

<u>In this figure</u> :

  • There are 6 carbon atoms
  • The Cl atom is bonded to the 2nd carbon atom

⇒ The Cl is a substituent group, termed as -chloro

⇒ Based on IUPAC nomenclature, the 6 atom chain starts with hex

⇒ There are only single bonds present, so it is an alkane

<u>The name is</u> :

  • <u>2-chlorohexane</u>
6 0
2 years ago
What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure? As air
Marina86 [1]

We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.

The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.

For any gaseous substance, density of gas is directly proportional to pressure of gas.

This can be explained from idial gas edquation:

PV=nRT

PV=\frac{w}{M}RT [where, w= mass of substance, M=molar mass of substance]

PM=\frac{w}{V}RT

PM=dRT [where, d=density of thesubstance]

So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.

Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.

3 0
3 years ago
Part A
Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4)  = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84  kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG =  ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

3 0
3 years ago
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