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guapka [62]
2 years ago
15

The element antimony has two stable isotopes, antimony-121 with a mass of 120.90 amu and antimony-123 with a mass of 122.90 amu.

From the atomic weight of Sb = 121.76 one can conclude that: ________
antimony-123 has the highest percent natural abundance
most antimony atoms have a mass of 121.76 amu
antimony-121 has the highest percent natural abundance
both isotopes have the same percent natural abundance
Chemistry
1 answer:
NARA [144]2 years ago
8 0

Answer:

antimony-121 has the highest percent natural abundance

Explanation:

percent natural abundance;

121.76 = 120.90 x + 122.90 (1 - x)

121.76 = 120.90 x + 122.90 - 122.90x

121.76 = -2x + 122.90

121.76 - 122.90 = -2x

x= 121.76 - 122.90/ -2

x= 0.57

Where x and 1 - x refers to the relative abundance of each of the isotopes

Percent natural abundance of antimony-121 = 57 %

Percent natural abundance of antimony-123 = (1 - 0.57) = 43%

Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.

We can clearly see that antimony-121 has the highest percent natural abundance.

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How many grams of precipitate will be formed when 20.5 mL of 0.800 M
Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

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Pachacha [2.7K]

Answer:

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Explanation:

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jolli1 [7]

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