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Jobisdone [24]
3 years ago
5

What would be the resulting molarity of a solution made by dissolving 17.8 grams of LiF in enough water to make a 915-milliliter

solution? Show all of the work needed to solve this problem.
Chemistry
1 answer:
wariber [46]3 years ago
3 0
First, determine the amount of substance in moles by dividing the given amount in grams by the molar mass. 
              17.8 grams LiF x (1 mole / 25.94 grams LiF) = 0.686 moles LiF
Divide this amount of substance by the volume of the solution in liters.
                           molarity = 0.686 moles LiF / 0.915 L sol'n
The molarity is therefore, 0.75 M. 
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Which of the following chemical formulas represents a common acid?
wariber [46]
HCI is one of the most common acids out of the following
5 0
3 years ago
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
Find the moles in 0.0023 of Co, at STP:
Alexus [3.1K]
What is does that mean 0.0023 cooooo
7 0
2 years ago
How to round the decimal 4.15 to the hundredths place?​
gladu [14]

Answer:4.15 is already in the hundreths place so you are good!

if it was 4.155 then you would round it to 4.16 but its NOT that so the answer IS 4.15

Explanation:

8 0
2 years ago
if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar
fgiga [73]

Answer:

A beaker  

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

6 0
2 years ago
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