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Alja [10]
3 years ago
10

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

he percent abundance of each isotope? Show all calculations and report to the correct number of sig figs.
Chemistry
1 answer:
Dvinal [7]3 years ago
6 0

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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Predict the sign of ΔSsys for each of the following processes. Which are positive? (a) Gasoline vapors mix with air in a car eng
OLEGan [10]

Answer:

Positive: a and b

Negative: c

Explanation:

The entropy (S) is the measure of the randomness of the system, and it intends to increase. The randomness can be determined by the energy of the molecules, their velocity and how distance they are between the other molecules.

When the entropy increases, ΔS is positive, when the entropy decreases, ΔS is negative. So, when gasoline mix with air in a car engine, the process intends to continue, the randomness increases and ΔS is positive. When hot air expands, the distance between the molecules increases, so ΔS is positive.

But, when humidity condenses, the molecules stay closer, so there's a decrease in the randomness, then ΔS is negative.

6 0
3 years ago
Sodium acetate can be formed from the metathesis/double replacement reaction of sodium
telo118 [61]

Answer:

Explanation:

Sodium Acetate Trihydrate BP Specifications

Sodium Acetate BP

C2H3NaO2,3H2O

Action and use

Used in solutions for dialysis; excipient.

DEFINITION

Sodium ethanoate trihydrate.

Content

99.0 per cent to 101.0 per cent (dried substance).

CHARACTERS

Appearance

Colourless crystals.

Solubility

Very soluble in water, soluble in ethanol (96 per cent).

IDENTIFICATION

A. 1 ml of solution S (see Tests) gives reaction (b) of acetates.

B. 1 ml of solution S gives reaction (a) of sodium.

C. Loss on drying (As shown in the Relevant Test).

TESTS

Solution S

Dissolve 10.0 g in carbon dioxide-free water prepared from distilled water R and dilute to 100 ml 100 ml with the same solvent.

Appearance of solution

Solution S is clear and colourless.

pH

7.5 to 9.0.

Dilute 5 ml of solution S to 10 ml with carbon dioxide-free water.

Reducing substances

Dissolve 5.0 g in 50 ml of water, then add 5 ml of dilute sulphuric acid and 0.5 ml of 0.002 M potassium permanganate. The pink colour persists for at least 1 h. Prepare a blank in the same manner but without the substance to be examined.

Chlorides

Maximum 200 ppm.

Sulphates

Maximum 200 ppm.

Aluminium

Maximum 0.2 ppm, if intended for use in the manufacture of dialysis solutions.

Arsenic

Maximum 2 ppm, determined on 0.5 g.

Calcium and magnesium

Maximum 50 ppm, calculated as Ca.

Heavy metals

Maximum 10 ppm.

Iron

Maximum 10 ppm, determined on 10 ml of solution S.

Loss on drying

39.0 per cent to 40.5 per cent, determined on 1.000 g by drying in an oven at 130C.

Sodium Acetate FCC Food Grade, US Food Chemical Codex

C2H3NaO2 Formula wt, anhydrous 82.03

C2H3NaO2·3H2O Formula wt, trihydrate 136.08

DESCRIPTION

Sodium Acetate occurs as colorless, transparent crystals or as a granular, crystalline or white powder. The anhydrous form is hygroscopic; the trihydrate effloresces in warm, dry air. One gram of the anhydrous form dissolves in about 2 mL of water; 1 g of the trihydrate dissolves in about 0.8 mL of water and in about 19 mL of alcohol.

Function: Buffer.

REQUIREMENTS

Identification: A 1:20 aqueous solution gives positive tests for Sodium and for Acetate.

Assay: Not less than 99.0% and not more than 101.0% of C2H3NaO2 after drying.

Alkalinity Anhydrous: Not more than 0.2%; Trihydrate: Not more than 0.05%.

Lead: Not more than 2 mg/kg.

Loss on Drying: Anhydrous: Not more than 1.0%; Trihydrate: Between 36.0% and 41.0%.

Potassium Compounds: Passes test.

5 0
3 years ago
If a gas had a volume of 6.7 L and started at STP, what would the new pressure be
Phantasy [73]

The new pressure would be = 4.46 atm

<h3>Further explanation</h3>

Given

V₁=6.7 L(at STP, 1 atm 273 K)

V₂=1.5 L

Required

The new pressure

Solution

Boyle's Law  

At a constant temperature, the gas volume is inversely proportional to the pressure applied  

\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}

P₂ = (P₁V₁)/V₂

P₂ = (1 atm x 6.7 L)/1.5 L

P₂ = 4.46 atm

5 0
2 years ago
What is a Metonic cycle? How does this influence when we see full moons?
vivado [14]

Answer:    A tropical year is longer than 12 lunar months and shorter than 13 of them. The arithmetical equation

12x12 + 7x13 = 235

allows it to be seen that a combination of 12 'shorter' (12 months) years and 7 'longer' (13 months) years will be equal to 19=12+7 solar years.

Explanation:

6 0
3 years ago
The half life of Radon-222 is 3.8 days. If you have a 90.0g sample of Radon-222 in the lab, how much will be left after 19 days?
statuscvo [17]

Answer:

half lives passed=5

given sample=90g

sample left=2.8125g

Explanation:

no. of half lives=total time/half life

no.=19days/3.8days

no.=5 days

after 5 half lives sample left=2.8125g

5 0
3 years ago
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