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Brums [2.3K]
3 years ago
7

How many liters of N2 gas are in 2.4 moles at STP?

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
7 0

Answer:

any gas takes up 22.4L per mole so 2.4*22.4=53.76

Explanation:

You might be interested in
La formula estructural del eter butil pentilico
VladimirAG [237]

Answer:

do you mean?

Explanation:

the structural formula of butyl pentyl ether

4 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
43. A stock glucose standard has a concentration of 1,000 mg/dL. A 1/5 dilution of this standard is made. What would be the fina
Nat2105 [25]

The final concentration of the diluted standard is 0.2 mg/dL.

<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>

Using the dilution formula:

  • C1V1 = C2V2

where

  • C1 is initial concentration
  • V1 initial volume
  • C2 is final concentration
  • V2 is final volume.

Assuming a final volume of 100 mL, and since a 1/5 dilution is made:

C1 = 1.00 mg/dL

V1 = 20

C2 = ?

V2 = 100 mL

C2 = C1V1/V2

C2 = 20 × 1/100

C2 = 0.2 mg/dL

Therefore, the final concentration of the diluted standard is 0.2 mg/dL.

Learn more about dilution at: brainly.com/question/24881505

6 0
2 years ago
Which of the following is likely to have the lowest viscosity? hot oil room temperature oil room temperature water below room te
zzz [600]

Viscosity, is a state of matter where it resists flow or may be in thick semifluid form because of friction acting inside the matter.In this case, hot oil room temperature would lessen the viscosity of the oil (animal oil). Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
4 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP! AND PLEASE DO NOT GIVE ME ANY VIRUSES!
ludmilkaskok [199]

Answer:

Hope it helps you.....

The answer is in form of image..

See it...

8 0
3 years ago
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