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Bas_tet [7]
2 years ago
6

A 2.10 L vessel contains 4.65 mol of nitrous oxide (N2O) at 3.82 atm. What will be the pressure of this gas in a container that’

s half the size?
Chemistry
1 answer:
sveticcg [70]2 years ago
4 0

Answer:

The answer to your question is 7.64 atm

Explanation:

Data

Volume 1 = V1 = 2.1 l

moles = 4.65

Pressure 1 = P1 = 3.82 atm

Volume 2 = Volume 1/2

Pressure 2 = ?

Process

1.- Calculate the new volume

Volume 2 = 2.10/2

Volume 2 = 1.05 l

2.- Use Boyle's law to find the Pressure

             P1V1 = P2V2

-Solve for P2

              P2 = P1V1 / V2

-Substitution

              P2 = (3.82)(2.1) / 1.05

-Simplification

              P2 = 8.022/1.05

-Result

              P2 = 7.64 atm

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Force = Mass * Acceleration = 59 kg * 9.75 m/s^2 = 575.25 N
6 0
3 years ago
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Starting from a 1 M stock of NaCl, how will you make 50 ml of 0.15 M NaCl?
WARRIOR [948]

Answer:

We need 7.5 mL of the 1M stock of NaCl

Explanation:

Data given:

Stock = 1M this means 1 mol/ L

A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL

Step 2: Calculate the volume of stock we need

The moles of solute will be constant

and n = M*V  

M1*V1 = M2*V2

⇒ with M1 = the initial molair concentration = 1M

⇒ with V1 = the volume we need of the stock

⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L

⇒ with M2 = the concentration of the new solution = 0.15 M

1*V1 = 0.15*(50)

V1 = 7.5 mL

Since 0.0075 L of 1M solution contains 0.0075 moles

50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M

We need 7.5 mL of the 1M stock of NaCl

6 0
2 years ago
Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t
Dvinal [7]

Answer :  The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

6.941=[(6.01512\times x)+(7.01600\times (100-x))]

By solving the term 'x', we get:

x=694.048

Percent abundance of Li isotope-1 = \frac{694.048}{100}=6.94\%

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

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