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3 years ago

If I have one mole of gold atoms, how many gold atoms do I have?

Chemistry

2 answers:

Bad White [126]3 years ago

5 0

Answer:

I hope it's helpful for you....

Explanation:

There are 6.022 X 1023 atoms in one mole of gold.

yuradex [85]3 years ago

3 0

Answer:

6.022 X 10^23

Explanation:

avatars number states that for every mole of an element there are 6.p22 X 10^23 atoms of that element

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How many moles is 7.78 x 1024<br> formula units of MgCl2?

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How many moles is 7.78 x 1024

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3 years ago

__________ is a characteristic of an organism that allows it to survive better in an environment.

Reil [10]

Adaptation is the characteristic of an organism that allows it to survive better in an environment

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3 years ago

2NaHCO3-->Na2CO3+H2O+CO2

algol [13]

I’m not sure what kind of answer you’re looking for, but if you’re supposed to translate it into a sentence it would be two moles of sodium bicarbonate(baking soda) decompose unto one mole of sodium carbonate(soda ash) plus one mole of water and one mole of carbon monoxide gas

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3 years ago

Since vinegar is 5% acetic acid and 95% water, find the mole ratio of acetic acid to water in 100 g of vinegar? Hint: Instead of

Oksanka [162]

The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.

<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>

The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.

The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.

In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.

Moles = mass/molar mass

molar mass of acetic acid = 62 g/mol

molar mass of water = 18 g/mol

moles of vinegar = 5/62 = 0.08

moles of water = 95/18 = 5.28

total moles = 5.36

Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36

Mole ratio of vinegar to water = 0.015 : 0.985

In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.

Learn more about mole ratio at: brainly.com/question/19099163

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7 0

2 years ago

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)

5 0

3 years ago