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Rzqust [24]
3 years ago
13

Which of the following changes will likely increase the solubility of a solid solute but decrease the solubility of a gaseous so

lute? A. Increasing the molecular mass of the solute B. Increasing the temperature C. Decreasing the concentration D. Decreasing the atmospheric pressure
Chemistry
2 answers:
Readme [11.4K]3 years ago
6 0
A. Increasing the molecular mass of the solute, Because that would Increase the solubility of solid solute, but wouldn't increase the gaseous solute.
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MA_775_DIABLO [31]3 years ago
5 0

Answer: Option (B) is the correct answer.

Explanation:

When we increase the temperature then there is increase in solubility of a solid because of more number of collisions between its particles. As a result, there will be an increase in rate of reaction.

But when we increase the temperature for a gaseous solute then its molecules will gain more kinetic energy due to which they collide more rapidly. Hence, they will escape out of the reaction mixture so, there will be decrease in the solubility of a gaseous solute.

Thus, we can conclude that increasing the temperature will most likely increase the solubility of a solid solute but decrease the solubility of a gaseous solute.

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An electron that is in the highest energy level of an atom and determines the atom's chemical properties is called a(n)
ss7ja [257]
The answer is C: Valence electron
3 0
3 years ago
If you know the mass of Earth, what other single value would you need to know to calculate Earth's average density?
Dennis_Churaev [7]

Answer:

Volume

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.  In order to find the density of earth when mass is given we have to calculate its volume. The volume of earth is calculated by using the volume formula for sphere. i.e 4/3 π r³. we also require radius to find the volume and we know that

Diameter = 2 × radius

The diameter of earth equator is 12756.75 Km. So we calculate the radius by dividing the diameter by 2. Then by putting the value of radius in 4/3 π r³ we will get the volume and then we can calculate the density of earth.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

7 0
3 years ago
What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?
Solnce55 [7]
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+  + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
     = -㏒(4.81x10^-7) = 6.32

3 0
3 years ago
Using the ideal gas law, PV=nRT, where R=0.0821 L atm/mol K, calculate the volume in liters of oxygen produced by the catalytic
AfilCa [17]

Answer:

V = 2.32 Liters

Explanation:

PV = nRT => V = nRT/P

n = 25.8g/122g/mole = 0.21 mole

R = 0.08206 L·atm/mol·K

T = 25.44°C + 273 = 298.44K

P = 2.22 atm (given in problem)

V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm

8 0
3 years ago
Calculate the mass of 3.5 mol C6H6
Lilit [14]
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
4 0
3 years ago
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