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lana66690 [7]
3 years ago
9

A gas occupies 56 L at 73°C. What volume will the gas occupy if the temp. cools to 0°C?

Chemistry
1 answer:
vova2212 [387]3 years ago
7 0

Answer:

44.2 L

Explanation:

Use Charles Law:

\frac{V1}{T1} =\frac{V2}{T2}

We have all the values except for V₂; this is what we're solving for. Input the values:

\frac{56 L}{346K} =\frac{V2}{273K}   -  make sure that your temperature is in Kelvin

From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:

\frac{56*273}{346} = V2

Therefore, V₂ = 44.2 L

It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.

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2 years ago
What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of table salt, NaCl?
tia_tia [17]

Answer:

3 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v           (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = 0.05 L

Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of Na and Cl.

Na= 22.99 g/mol         Cl= 35.45g/mol

2.) Add them together to have their combined molecular weights.

22.99 + 35.45= 58.44 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 58.44 g = 0.17111 mol

4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.17111 mol / 0.05 L = 3.4222

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 3.4222 to 3 mol/L

Sorry this explanation is long let me know if you need a better more written out sample.

3 0
3 years ago
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