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zvonat [6]
3 years ago
7

What would happen to the results if the plastic wrap was dirty in a water filtration experiment?​

Chemistry
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:

The plastic wrap of the covered cup acts like the atmosphere, and traps the water vapor. In a real cloud, the water vapor cools back into liquid water. In the covered cup, the air can only hold so much vapor, and the vapor condenses back to liquid water forming a “rain cloud” on the plastic wrap.

Explanation:

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The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure
AlekseyPX

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1 = initial pressure = 198 kPa

P_2 = final pressure = 225 kPa

T_1 = initial temperature = 27^oC=273+27=300K

T_2 = final temperature = ?

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

\frac{198kPa}{300K}=\frac{225kPa}{T_2}

T_2=340.9K\approx 341K

Therefore, the temperature of the air in the tire is, 341 K

5 0
3 years ago
The molarity of a solution prepared by dissolving 4.11 g of NaI in enough water to prepare 312 mL of solution is
Dimas [21]

Answer:

The correct answer is 8.79 × 10⁻² M.

Explanation:

Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,

No. of moles of NaI = Weight of NaI/ Molecular mass

= 4.11 / 149.89

= 0.027420

The vol. of the solution given is 312 ml or 0.312 L

The molarity can be determined by using the formula,

Molarity = No. of moles/ Volume of the solution in L

= 0.027420/0.312

= 0.0879 M or 8.79 × 10⁻² M

6 0
2 years ago
Which of the following statements is true about one formula unit of RuF2?<br>​
valina [46]
For me the answer is E. It is composed of one Ru2+ ion and two F- ions.
6 0
3 years ago
What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of
Elina [12.6K]
Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

<span> ΔT(freezing point)  = (Kf)mi
3  = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>

4 0
3 years ago
Read 2 more answers
An aqueous solution contains
ivolga24 [154]

Answer:

32.7

Explanation:

I just did it and got it right

4 0
2 years ago
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