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2 years ago

B. from 52 moles of nitric acid, how many moles of silver nitrate will be produced?

Chemistry

1 answer:

IceJOKER [234]2 years ago

6 0

Answer:

39 mol AgNO3

Explanation:

We have the equation 4HNO3 + 3Ag -----> 3AgNO3 + NO + 2H2O

We want to calculate the number of silver nitrate (AgNO3) moles that would be produced from 52 moles of nitric acid ( HNO3 )

We can calculate this by using mole ratio as well as dimensional analysis.

The mole ratio of Silver nitrate to nitric acid based on the balanced equation is 3AgNO3:4HNO3.

Using this we can create a table: The table is attached.

Breakdown of the table.

The moles of nitric acid cancel out and we multiply 52 by 3/4 to get 39 moles of Silver nitrate.

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Magnesium has three naturally occurring isotopes(Mg-24, Mg-25, and Mg-26). The atomic mass and natural abundance of Mg-24 are 23

xxMikexx [17]

Answer:

The answer to your question is below

Explanation:

Data

Atomic mass Abundance

Mg -24 = 23.9850 79%

Mg -25 = 24.9858 10%

Mg - 26= ? ?

Process

1.- Find the abundance of Mg-26

The sum of the abundance of the three isotopes equals 100%

Abundance Mg-24 + Abundance-Mg-25 + Abundance-26 = 100

79 + 10 + Abundance-26 = 100

Abundance Mg-26 = 100 -89

Abundance Mg-26 = 11%

2.- Find the atomic mass of Mg-26, the average atomic mass is 24.31 amu

Average atomic mass = (abundance)(atomic mass- 24) + (abundance)

(atomic mass-25) + (abundance)(atomic mass-26)

24.31 = (0.79)(23.9850)+ (0.10)(24.9858) + (Atomic mass-26)(0.11)

24.31 = 18.94815 + 2.4986 + 0.11(atomic mass-26)

24.31 = 21.4342 + 0.11(atomic mass- 26)

0.11(atomic mass-26) = 24.31 - 21.4342

0.11(atomic mass-26) = 2.8758

Atomic mass-26 = 2.8758/0.11

Atomic mass-26 = 26.1436 amu

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2 years ago

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Mr. Bullis wants to know if he gives teachers candy, it will make them smile more. He gives half the teachers candy and the othe

mote1985 [20]

Answer:

the experimental group would be the teachers who recieved the candy

control group would be the teachers who dont get any candy

Explanation:

experimental group is the group that receives the test variable being tested in this case the variable being the candy

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purpose of having a control is to rule out other factors which may influence the results of an experiment.

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According to equation at STP,

67.2 L (3 moles) of H₂O is produced by = 78 g of Al(OH)₃
So,
65.0 L of H₂O will be produced by = X g of Al(OH)₃

Solving for X,
X = (65.0 L × 78 g) ÷ 67.2 L

X = 75.44 g of Al(OH)₂
Result:
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