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Nostrana [21]
3 years ago
10

If the atmospheric pressure in the laboratory is 1.2 atm, how many moles of gas were in each syringe? (Hint: Choose one volume a

nd temperature pair from your data table to use in your ideal gas law calculation.)

Chemistry
1 answer:
Naily [24]3 years ago
8 0

Answer:

A: 2.525 x 10-4 mol

B: 2.583 x 10-4 mol

Explanation:

Part A:

Data Given:

. Temperature of water (H2O) = 21.3°C

Convert Temperature to Kelvin

T = °C + 273

T = 21.3 + 273 = 294.3 K

volume of (H2O) gaseous state = 5.1 mL

Convert mL to liter

1000 mL = 1L

5.1 ml = 5.1/1000 = 0.0051 L

Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n=PV/RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/ mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0051 L / 0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0061 atm.L / 24.162 L.atm.mol-1

n = 2.525 x 10-4 mol

no. of moles of gas (H2O) = 2.525 x 10-4 mol

Part B:

Data Given:

Temperature of water (H2) = 21.3°C

Convert Temperature to Kelvin

T = "C + 273

T= 21.3 + 273 = 294.3 K

volume of (H2) gas = 5.2 mL

Convert mL to liter

1000 mL = 1 L

5.2 ml = 5.2/1000 = 0.0052 L

Pressure = 1.2 atm

. no. of moles = ?

Solution

no. of moles can be calculated by using ideal gas formula

PV = nRT

Rearrange the equation for no. of moles

n= PV / RT......... (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

where

R = 0.08206 L.atm/mol. K

Now put the value in formula (1) to calculate no. of moles of

n = 1.2 atm x 0.0052 L/0.08206 L.atm.mol-1. K-1 x 294.3 K

n = 0.0062 atm.L/ 24.162 L.atm.mol-1

n = 2.583 x 10-4 mol

I

no. of moles of gas (H2) = 2.583 x 10-4 mol

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Starting with 6.3 g of salicylic acid, how many grams of acetylsalicylic acid (theoretical yield) can be made? Assume salicylic
tensa zangetsu [6.8K]

Answer:

We will produce 8.2 grams of acetylsalicylic acid

Explanation:

step 1: Data given

Mass of salicylic acid, = 6.3 grams

Molar mass salicylic acid = 138.12 g/mol

Molar mass of acetylsalicylic acid = 180.158 g/mol

Step 2: The balanced equation

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

Step 3: Calculate moles salicylic acid

Moles salicylic acid = mass salicylic acid / molar mass salicylic acid

Moles salicylic acid = 6.3 grams / 138.12 g/mol

Moles salicylic acid = 0.0456 moles

Step 4: Calculate moles acetylsalicylic acid

Since the mole ratio is 1 to 1

For 0.0456 moles salicylic acid we'll have 0.0456 moles acetylsalicylic acid

Step 5: Calculate mass acetylsalicylic acid

MAss acetylsalicylic acid = moles acetylsalicylic acid * molar mass acetylsalicylic acid

MAss acetylsalicylic acid = 0.0456 moles * 180.158 g/mol

Mass acetylsalicylic acid = 8.2 grams

We will produce 8.2 grams of acetylsalicylic acid

4 0
3 years ago
How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
strojnjashka [21]
2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
8 0
3 years ago
What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH4Cl produced by the reaction
Mashcka [7]

Answer:

a) HCl is the limiting reagent.

b) Mass of NH₄Cl formed = 6.68 g

c) Pressure of the gas remaining in the flask = 1.742 atm

Explanation:

The complete Question is presented in the attached image to this solution.

To solve this question, we first need to obtain the limiting regaent for this reaction.

The limiting reagent is the reagent that is in short supply in the reaction and is used up in the reaction. It determines the amount of products that will be formed and the amount of other reactants that will be required for the reaction.

NH₃ (g) + HCl (g) ⟶ NH₄Cl (s)

1 mole of NH₃ reacts with 1 mole of HCl

we first convert the masses of the gases available to number of moles.

Number of moles = (Mass/Molar Mass)

Molar mass of NH₃ = 17.031 g/mol, Molar mass of HCl = 36.46 g/mol

Number of moles of NH₃ = (4.55/17.031) = 0.2672 mole

Number of moles of HCl = (4.55/36.46) = 0.1248 mole

Since 1 mole of NH₃ reacts with 1 mole of HCl

It is evident that HCl is in short supply and is the limiting reagent.

NH₃ is in excess.

So, to calculate the amount of NH₄Cl formed,

1 mole of HCl gives 1 mole of NH₄Cl

0.1248 mole of HCl will also gove 0.1248 mole of NH₄Cl

Mass (Number of moles) × (Molar Mass)

Molar mass of NH₄Cl = 53.491 g/mol

Mass of NH₄Cl formed = 0.1248 × 53.491 = 6.68 g

c) The gas remaining in the flask is NH₃

0.1248 mole of NH₃ is used up for the reaction, but 0.2672 mole was initially available for reaction,

The amount of NH₃ left in the reacting flask is then

0.2672 - 0.1248 = 0.1424 mole.

Using the ideal gas Equation, PV = nRT

We can obtain the rrequired pressure of the remaining gas in the flask

P = Pressure = ?

V = Volume = 2.00 L

n = number of moles = 0.1424 mole

R = molar gas constant = 0.08205 L.atm/mol.K

T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K

P = (nRT/V)

P = (0.1424×0.08205×298.15/2) = 1.742 atm

Hope this Helps!!!

7 0
3 years ago
Need help please :'((((((((((
kati45 [8]

Answer:

you work out the moles by finding the grand formula mass them you divide the actual mass by the grand formula mass: 25/160 (my grand formula mass is probably slightly different from yours) which I got 0.15625 then I times it by 32(O2) and I got 5. but my periodic table is different. it's whole numbers. ;-)

Explanation:

4.44

6 0
2 years ago
Plz help. Look at the image shown answer 1.2.3 please
horsena [70]
Re enema ughh ten Hessian
6 0
3 years ago
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