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gulaghasi [49]
2 years ago
7

Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative

abundance of 7.49%.
What is the abundance of the other isotope?
What is the mass of the other isotope?
Chemistry
1 answer:
koban [17]2 years ago
3 0

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

6.941 = (6.015×7.49)+(x×92.51) /100

6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

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Answer:

1.98 M

Explanation:

Given data

  • Initial volume (V₁): 93.2 mL
  • Initial concentration (C₁): 2.03 M
  • Volume of water added: 3.92 L

Step 1: Convert V₁ to liters

We will use the relationship 1 L = 1000 mL.

93.2mL \times \frac{1L}{1000mL} = 0.0932 L

Step 2: Calculate the final volume (V₂)

The final volume is the sum of the initial volume and the volume of water.

V_2 = 0.0932L + 3.92 L = 4.01L

Step 3: Calculate the final concentration (C₂)

We will use the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.03 M \times 3.92L}{4.01L} = 1.98 M

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2 years ago
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LUCKY_DIMON [66]
You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25

That should be the right answer but I'm not sure. It has been awhile since I have done this.

4 0
1 year ago
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What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?
Alex

Answer:

\large \boxed{93\, \% }

Explanation:

1. Calculate the molar mass of N₂H₄

2N = 2 × 14 = 28

4H = 2 ×  1  = <u>  4</u>

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\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}

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kati45 [8]
First step is to get the mass of the mercury:
Pressure = mass/volume
mass = pressure x volume = 13.5 x 1.85 = 24.975 gm

Second step is to calculate the number of moles in 24.975 gm:
From the periodic table, the molar mass of mercury is 200.59
mass = number of moles x molar mass
number of moles = 24.975 / 200.59 = 0.1245 mole

Last step is to get the number of atoms:
Each mole contains 6.02 x 10^23 atoms
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maw [93]
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Remember that molarity is moles per liter:
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8 0
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