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mr_godi [17]
4 years ago
15

A sealed, expandable container is initially at 2.000 L and 25°C. To what temperature must it be changed to have a volume of 6.00

L?
..I used V1/T1 = V2/T2 to get 75 degrees Celsius, but for some reason that's wrong. I'd love to know what I did wrong.
Chemistry
1 answer:
balandron [24]4 years ago
4 0

Answer:

The new temperature is 894 K or 621 °C

Explanation:

Step 1: Data given

Initial volume of the container = 2.000L

Initial temperature = 25.0 °C = 298 K

Volume increased to 6.00 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 2.00L

⇒with T1 = the initial temperature = 25 °C = 298 K

⇒with V2 = the increased volume 6.00 L

⇒with T2 = the new temperature

2.00 L / 298 K = 6.00 L / T2

T2 = 894 K = 621 °C

The new temperature is 894 K or 621 °C

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givi [52]

Answer: Option (c) is the correct answer.

Explanation:

A limiting reagent is defined as a reagent that completely gets consumed in a chemical reaction. A limiting reagent limits the formation of products.

For example, we have given 5 mol of A and the reaction is 2A + 4B \rightarrow 2AB

Whereas when 4 mol B will react with 2 mol of A. Hence, 8 mol of B will react with 4 mol A as follows.

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Thus, we can conclude that limiting reactant is the term used to describe the reactant that is used up completely and controls the amount of product that can be produced during a chemical reaction.  

6 0
4 years ago
How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm?
meriva
Hello!

Data:

P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)

Apply the data to the Clapeyron equation (ideal gas equation), see:
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1*18.5 = n*0.082*300

18.5 = 24.6n

24.6n = 18.5

n =  \dfrac{18.5}{24.6}

\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark

Note: If the feedback is to be considered, the closest response is 0.751 mol Nacl

_________________
_________________

I hope this helps. =)


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