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avanturin [10]
3 years ago
15

How many mL are 2.3 mol of CO2 at STP?

Chemistry
2 answers:
kupik [55]3 years ago
8 0

Answer:

0.052mL

Explanation:

1mole of a gas occupy 22.4L.

Therefore, 1 mole of CO2 will also occupy 22.4L.

If 1mole of CO2 occupies 22.4L,

Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L

coverting this volume to mL, we simply divide by 1000 as shown below:

51.52/1000 = 0.05152mL = 0.052mL

Dovator [93]3 years ago
7 0

Answer:

The volume of 2.3 moles CO2 is 51.5 L = 51500 mL

Explanation:

Step 1: Data given

Number of moles CO2 = 2.3 moles

Step 2: Calculate volume CO2

p*V= n*R*T

⇒with p = the pressure = 1 atm

⇒with V = the volume = TO BE DETERMINED

⇒with n = the number of moles = 2.3 moles

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 273 K

V = (nRT)/p

V = (2.3*0.08206*273)/1

V = 51.5 L

The volume of 2.3 moles CO2 is 51.5 L = 51500 mL

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A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is
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At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

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M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-

Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

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