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2 years ago

1) How many aluminum atoms are there in 3.50 grams of Al2O3?

Chemistry

1 answer:

drek231 [11]2 years ago

3 0

Answer:

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

Step 1: Data given

Mass of Al2O3 = 3.50 grams

Molar mass of Al2O3 = 101.96 g/mol

Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

Moles Al2O3 = 0.0343 moles

Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al

Step 4: Calculate aluminium atoms

Aluminium atoms = moles aluminium * Number of Avogadro

Aluminium atoms = 0.0686 * 6.022 * 10^23

Aluminium atoms = 4.13 *10^22 aluminium atoms

The correct answer is E

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How many moles of AgNO3 must react to form 0.854 mol Ag?

kipiarov [429]

The answer:
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and </span>AgNO3 → Ag + (aq) + NO3- (aq)
1 mol 1mol 1mol
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2 years ago

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Elis [28]

The concentration of the drug stock solution is 1.5*10^-9 M i.e. 1.5 * 10^-9 moles of the drug per Liter of the solution

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1 mole of the drug will contain 6.023*10^23 drug molecules

Therefore, 1.5*10^-12 moles of the drug will correspond to :

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The number of cancer cells = 2.0 * 10^5

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2 years ago

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A student wants to reclaim the iron from an 18.0-gram sample of iron(III) oxide, which

lilavasa [31]

Answer:

$m_{Fe}=12.6gFe$

Explanation:

Hello,

In this case, since we have grams of iron (III) oxide whose molar mass is 159.69 g/mol are able to compute the produced grams of iron by using its atomic mass that is 55.845 g/mol and their 2:4 molar ratio in the chemical reaction:

$m_{Fe}=18.0gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_3}*\frac{4molFe_2O_3}{2molFe_2O_3} *\frac{55.845gFe}{1molFe_2O_3} \\\\m_{Fe}=12.6gFe$

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2 years ago

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2.1 liters is the same as cm3 and ml

Kipish [7]

1L = 1000ml
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2 years ago

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creativ13 [48]

Answer:

O FX will be greater than FY

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