Please help I’m stuck

Determine the formula weights of each of the following compounds.Part A) Nitrous oxide, N2O, known as laughing gas and used as a

densk [106]

Answer:

See explanation

Explanation:

a) Nitrous oxide (N2O) has a molar mass of 44.014 amu. It has 2 nitrogen atoms each with a mass of 14.007 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 44.014 amu ) *100% = 63.6%

Percentage oxygen = (16 amu/44.014 amu) * 100% = 36.4 %

63.6% + 36.4% = 100%

b) Benzoic acid (C7H6O2) has a molar mass of 122.13 amu. It has 6 hydrogen atoms each with a mass of 1.01 amu; it has 7 carbon atoms each with a mass of 12.01amu and 2 oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (6*1.01 amu / 122.13 amu)*100% = 4.96%

Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

Percentage oxygen = (2*16 amu/58.32 amu)*100% = 54.9%

d) Urea CO(NH2)2 has a molecular mass of 60.064 amu. It has 2 Nitrogen atoms each with a mass of 14.007 amu, 4 hydrogen atoms each with a mass of 1.01 amu,1 carbon atom with a mass of 12.01 amu and 1 oxygen atom with a mass of 16.0 amu.

Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

Percentage hydrogen = (4*1.01 amu/60.064amu)*100% = 6.72%

Percentage carbon = (12.01 amu/60.064amu)*100% = 20.0%

Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

e) Osopentyl acetate (C7H14O2) has a molecular mass of 130.2 amu. It has 14 hydrogen atoms each with a mass of 1.01 amu,7 carbon atoms each with a mass of 12.01 amu and 2 oxygen atom with a mass of 16.0 amu.

Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

Percentage oxygen =(2*16 amu/130.2 amu)*100% = 24.6%

7 0

3 years ago

Calculate the volume in liters of a potassium iodide solution that contains of potassium iodide . Be sure your answer has the co

Anna11 [10]

Answer:

23.8 L

Explanation:

There is some info missing. I think this is the original question.

Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.

The molar mass of potassium iodide is 166.00 g/mol. The moles corresponding to 150 grams are:

150 g × (1 mol/166.00 g) = 0.904 mol

0.904 moles of potassium iodide are contained in an unknown volume of a 0.0380 mol/L potassium iodide solution. The volume is:

0.904 mol × (1 L/0.0380 mol) = 23.8 L

6 0

3 years ago

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

posledela

The question is incomplete, here is the complete question:

When silver nitrate reacts with copper, copper(II) nitrate and silver are produced. The balanced equation for this reaction is:

$2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag$

Suppose 6 moles of silver nitrate react. The reaction consumes___ moles of copper. The reaction produces __ moles of copper(II) nitrate and __ moles of silver.

Answer: The amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

Explanation:

We are given:

Moles of silver nitrate = 6 moles

For the given chemical reaction:

$2AgNO_3+Cu\rightarrow Cu(NO_3)_2+2Ag$

For copper metal:

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of copper metal

So, 6 moles of silver nitrate will react with = $\frac{1}{2}\times 6=3mol$ of copper metal

Moles of copper reacted = 3 moles

For copper(II) nitrate:

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 1 mole of copper(II) nitrate

So, 6 moles of silver nitrate will produce = $\frac{1}{2}\times 6=3mol$ of copper(II) nitrate

Moles of copper(II) nitrate produced = 3 moles

For silver metal:

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6 moles of silver nitrate will produce = $\frac{2}{2}\times 6=6mol$ of silver metal

Moles of silver metal produced = 3 moles

Hence, the amount of copper metal reacted is 3 moles, amount of copper (II) nitrate produced is 3 moles and amount of silver metal produced is 6 moles.

6 0

3 years ago

3. A mass of 0.15 ounces is equal to how many grams?

garik1379 [7]

Answer:

Option C = 4.25 g

Explanation:

Ounce and grams are unit of mass. Ounce is larger unit while gram is smaller unit. The one ounce is consist of 28.35 g or we can say that one ounce is equal to 28.35 g. In order to convert the given ounce value into grams the value is multiply with 28.35 g.

Given data:

Mass = 0.15 ounce

Mass in gram = ?

Solution:

One ounce is equal to 28.35 g, so

0.15 × 28.35 = 4.25 g

5 0

3 years ago

if it takes 54 ml of 0.100 M naoh to neutralize 125ml of an hcl solution, what is the concentration of the hcl

vichka [17]

Answer:

c = 0.0432moldm ^−3

Explanation:

The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:

Acid+Base ->Salt+ Water

4 0

2 years ago