All categories

3 years ago

A balloon has a volume of 3.0 L at room temperature (27oC). At what temperature would the balloon have a volume of 4.0L?

Chemistry

1 answer:

diamong [38]3 years ago

6 0

Answer: 400K

Explanation:

Given that,

Original volume of balloon V1 = 3.0L

Original temperature of balloonT1 = 27°C

Convert the temperature in Celsius to Kelvin

(27°C + 273 = 300K)

New volume of balloon V2 = 4.0L

New temperature of balloon T2 = ?

Since volume and temperature are given while pressure is constant, apply the formula for Charle's law

V1/T1 = V2/T2

3.0L/300K = 4.0L/T2

To get the value of T2, cross multiply

3.0L x T2 = 4.0L x 300K

3.0LT2 = 1200LK

Divide both sides by 3.0L

3.0LT2/3.0L = 1200LK/3.0L

T2 = 400K

Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.

You might be interested in

Is flammability an example of physical property

kipiarov [429]

It's a chemical property

5 0

3 years ago

Read 2 more answers

how would it affect your results if you use a wet Erlenmeyer flask instead of a dry one when transferring your acid solution fro

zloy xaker [14]

With a wet Erlenmeyer flask, the result will not be accurate. It might be contaminated by any type of liquid such as basic or acidic. Therefore, an error will occur since it is not 100% acid.- hope this helped

8 0

3 years ago

Read 2 more answers

Elements, Compounds,& Mixtures,

mrs_skeptik [129]

Answer:

1. Elements: F, H, E

2. Compounds: A, D, I

3. Mixtures: B, C, G

Explanation:

3 0

3 years ago

What is the specific latent heat of a fussion for a substance that takes 550 kj to melt 14 kg at 262 k?

Lina20 [59]

The latent heat is correlated with energy as follows:

Q = mL

550 * 103 = 14 * 103 * L

L = 39.285 J /g

Thus, latent heat of the substance is 39.285 j /g

7 0

3 years ago

What is the pH of this solution?

Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}$

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

$1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M$

Which is also:

$[OH^-]=1.643x10^{-3}M$

Thereafter we can compute the pOH first:

$pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784$

Finally, the pH turns out:

$pH=14-2.784\\\\pH=11.216$

Regards!

5 0

3 years ago