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faltersainse [42]
3 years ago
9

How many moles are there in 5.30 x 1024 atoms of silver?

Chemistry
1 answer:
Rus_ich [418]3 years ago
6 0

Answer:

The mass of

4.6

×

10

24

atoms of silver is approximately 820 g.

Explanation:

In order to determine the mass of a given number of atoms of an element, identify the equalities between moles of the element and atoms of the element, and between moles of the element and its molar mass.

1

mole atoms Ag=6.022xx10

23

atoms Ag

Molar mass of Ag =#"107.87 g/mol"#

Multiply the given atoms of silver by

1

mol Ag

6.022

×

23

atoms Ag

. Then multiply times the molar mass of silver.

4.6

×

10

24

atoms Ag

×

1

mol Ag

6.022

×

10

23

atoms Ag

×

107.87

g Ag

1

mol Ag

=

820 g Ag

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Hydrazine (N2H4) is used as rocket fuel. It reacts with oxygen to form nitrogen and water.
Marina86 [1]

Answer:

See explanation below for answers

Explanation:

This is a stochiometry reaction. LEt's write the overall reaction again:

N₂H₄ + O₂ ---------> N₂ + 2H₂O

This reaction is taking place at Standard temperature and pressure conditions (STP) which are P = 1 atm and T = 273 K.  To know the volume of N₂ formed, we need to know first how many moles are formed, and this can be calculated with the reagents and the limiting reagent. Let's calculate the moles first of the reagents:

MM N₂H₄ = 32 g/mol;    MM O₂ = 32 g/mol

mol N₂H₄ = 2000 / 32 = 62.5 moles

mol O₂ ? 2100 / 32 = 65.63 moles

Now that we have the moles, we need to apply the stochiometry and calculate the limiting reagent. According to the overall reaction we have a mole ratio of 1:1 between N₂H₄ and O₂, therefore:

1 mole N₂H₄ ---------> 1 mole O₂

62.5 moles ----------> X

X = 62.5 moles of O₂

But we have 65.63 moles, therefore, the limiting reactant is the N₂H₄.

We also have a 1:1 mole ratio with the N₂, so:

moles N₂H₄ = moles N₂ = 62.5 moles

Now that we have the moles, we can calculate the volume with the ideal gas equation:

PV = nRT

V = nRT / P

R: gas constant (0.082 L atm / K mol)

Replacing we have:

v = 62.5 * 0.082 * 273 / 1

V = 1399.13 L of N₂

Now, how many grams of the excess remains?, we know how many moles are reacting so, let's see how much is left:

moles remaining = 65.63 - 62.5 = 3.12 moles

then the mass of oxygen:

m = 3.12 * 32 = 100.16 g of O₂

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Rzqust [24]

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Why?

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0.00000000159 km^{2}=1.59*10^{-9} km^{2}

Have a nice day!

#LearnwithBrainly

8 0
3 years ago
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