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Alina [70]
3 years ago
13

Classify these molecules as polar or nonpolar. Drag each item to the appropriate bin.

Chemistry
1 answer:
mylen [45]3 years ago
8 0

Answer:

This queston does not provides the molecules that are needed to be checked in order to classify them as polar or non-polar.

Explanation:

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A rigid cylinder with a movable piston contains a sample of hydrogen gas. At 330. K, this sample has a pressure of 150. kPa and
Marta_Voda [28]

Answer:

V₂ = 4.34 L

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 3.50 L

Initial pressure = 150 Kpa (150/101.325 = 1.5 atm)

Initial temperature = 330 K

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ =  1.5 atm ×  3.50 L × 273 K / 330 K × 1 atm

V₂ = 1433.3 atm .L. K / 330 k.atm

V₂ = 4.34 L

4 0
3 years ago
9, What is an Independent
stepan [7]

Answer:

Not influenced or controlled by others in matters of opinion

8 0
3 years ago
A football is inflated to 12psi at 20°C What is the new pressure in the football
Radda [10]

Answer: The new pressure is 3 psi.

Explanation:

Given: P_{1} = 12 psi,         T_{1} = 20^{o}C

P_{2} = ?,          T_{2} = 5^{o}C

Formula used is as follows.

\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\frac{12 psi}{20^{o}C} = \frac{P_{2}}{5^{o}C}\\P_{2} = \frac{12 psi \times 5^{o}C}{20^{o}C}\\= 3 psi

Thus, we can conclude that the new pressure is 3 psi.

7 0
3 years ago
1.6x10^23 lead atoms. Find the weight in grams
Ber [7]
Moles of lead(Pb) = 1.6x10^23/6.02x10^23 = 0.265 moles.

Weight of lead = moles x atomic weight of lead
                         =  0.265x207.2
                         =  54.908 grams.

Hope this helps!
5 0
3 years ago
Why is it permissible to use a wet bottle when first obtaining your kmno4 solution?
Murrr4er [49]
For the answer to the question above, well presumably because the exact concentration of the composition KMnO4 solution doesn't matter. <span>If the concentration of the KMnO4 solution is important (usually in titrations etc.) then it is not allowed to use a wet bottle. The water in the bottle will dilute the KMnO4 solution and change the concentration of the said compound.</span>


5 0
3 years ago
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