Are these correct?

1. What is the formula mass for Mg(ClO3)2?
mg=24.31
cl=35.45 x 2 = 70.9
o=16 x 6 = 96
24.31+70.9+96= Answer:191.2 amu

2.How many molecules of CO2 are in 4.5 moles of CO2?
4.5 x 6.022 x10^23=Answer 2.7x10^24

3.How many grams are in 9.21 moles of H2CO3?
h=1.008 x2=2.016
c=12.01
o=16 x3=48
2.016+12.01+48=62.026g/mol
62.026 x 9.21=Answer 571g

4.How many moles are present in 88.2 grams of NaCl?
na=22.99
cl=35.45
22.99+35.45=58.44g/mol
88.2/58.44= Answer 1.51 mol

5. The percent composition of water is
A. 7% H, 33% O
B. 2 H, 1 O
C. 11% H, 89% O
D. 2 H, 16 O
E. 5.9%H, 94.1%O

water=h2o
h=1.008 x 2=2.016
0=16
2.016+16=18.016
2.016/18.016= 11%
16/18.016=89%
Answer= C. 11% H, 89% O

Thank you for the help. This has been a challenging chapter for me and I want to make sure I am doing it correctly.

Looks like #5 got cut off

water=h2o
h=1.008 x 2=2.016
0=16
2.016+16=18.016
2.016/18.016= 11%
16/18.016=89%
Answer= C. 11% H, 89% O

Thank you for the help. This has been a challenging chapter for me and I want to make sure I am doing it correctly.

All of these look ok to me.

Let's go through each question to verify if the answers are correct:

1. The formula mass for Mg(ClO3)2 is calculated correctly. The values for Mg, Cl, and O are accurate, and the sum of these values is 191.2 amu. Therefore, the answer is correct.

2. The calculation for the number of molecules of CO2 in 4.5 moles of CO2 is also correct. The value of Avogadro's number (6.022 x 10^23) is used correctly, and multiplying it by 4.5 moles yields 2.7 x 10^24 molecules of CO2. Therefore, the answer is correct.

3. The calculation for the number of grams in 9.21 moles of H2CO3 is mostly correct. The values for H, C, and O are accurate, and the sum of these values is approximately 62.03 g/mol (rounded to the hundredths place). However, when multiplying this value by 9.21 moles, the correct answer should be approximately 570.36 g (rounded to the hundredths place). Therefore, the answer should be 570 g, rather than 571 g.

4. The calculation for the number of moles present in 88.2 grams of NaCl is accurate. The values for Na and Cl are correct, and the sum of these values is 58.44 g/mol. Dividing 88.2 grams by 58.44 g/mol yields approximately 1.51 moles (rounded to the hundredths place). Therefore, the answer is correct.

5. The percent composition of water is calculated correctly. The values for H and O are accurate, and when divided by the molar mass of water (18.016 g/mol), they yield approximately 11% and 89%, respectively (rounded to the nearest whole number). Therefore, the answer is correct and matches option C.

Overall, you did a great job with your calculations. The only minor error was in question 3, where the answer should be 570 g instead of 571 g. Keep up the good work!

1. To find the formula mass for Mg(ClO3)2, you need to calculate the atomic masses of each element in the compound and then multiply by the respective number of atoms.

- Mg: The atomic mass of magnesium (Mg) is 24.31 amu.
- Cl: The atomic mass of chlorine (Cl) is 35.45 amu. However, there are 2 chlorine atoms in the compound, so you multiply by 2, which gives you 70.9 amu.
- O: The atomic mass of oxygen (O) is 16 amu. There are 3 oxygen atoms in the compound, so you multiply by 3, which gives you 48 amu.

Now, add up the atomic masses of each element:
24.31 + 70.9 + 48 = 143.21 amu

So, the formula mass for Mg(ClO3)2 is 143.21 amu.

2. To calculate the number of molecules in a given number of moles, you can use Avogadro's number, which represents the number of entities (atoms, molecules, etc.) in one mole of a substance. Avogadro's number is approximately 6.022 x 10^23.

In this case, you have 4.5 moles of CO2. To find the number of molecules, you multiply the number of moles by Avogadro's number:
4.5 moles x 6.022 x 10^23 molecules/mole = 2.7 x 10^24 molecules

So, there are approximately 2.7 x 10^24 molecules of CO2 in 4.5 moles.

3. To determine the mass of a given number of moles, you need to know the molar mass of the compound, which is the mass of one mole of the substance.

In this case, you have 9.21 moles of H2CO3. To calculate the mass, you need to find the molar mass of H2CO3:
- H: The atomic mass of hydrogen (H) is 1.008 amu. Since there are 2 hydrogen atoms in the compound, you multiply by 2, which gives you 2.016 amu.
- C: The atomic mass of carbon (C) is 12.01 amu.
- O: The atomic mass of oxygen (O) is 16 amu. There are 3 oxygen atoms in the compound, so you multiply by 3, which gives you 48 amu.

Now, add up the atomic masses of each element:
2.016 + 12.01 + 48 = 62.026 g/mol

Finally, multiply the molar mass by the number of moles:
62.026 g/mol x 9.21 moles = 571 g

So, there are 571 grams in 9.21 moles of H2CO3.

4. To find the number of moles in a given mass, you need to divide the mass by the molar mass of the substance.

In this case, you have 88.2 grams of NaCl. The molar mass of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl):
- Na: The atomic mass of sodium (Na) is 22.99 amu.
- Cl: The atomic mass of chlorine (Cl) is 35.45 amu.

Now, add up the atomic masses of each element:
22.99 + 35.45 = 58.44 g/mol

Divide the mass by the molar mass to find the number of moles:
88.2 g / 58.44 g/mol = 1.51 mol

So, there are approximately 1.51 moles in 88.2 grams of NaCl.

5. To find the percent composition of a compound, you need to determine the mass of each element and then calculate the percentage.

In this case, the compound is water (H2O). You need to find the mass of hydrogen (H) and oxygen (O) in water.

- H: The atomic mass of hydrogen (H) is 1.008 amu. There are 2 hydrogen atoms in water, so multiply by 2, which gives you 2.016 amu.
- O: The atomic mass of oxygen (O) is 16 amu.

Add up the atomic masses of each element:
2.016 + 16 = 18.016 g/mol

To find the percent composition, divide the mass of each element by the molar mass and multiply by 100:
- For hydrogen (H): (2.016 g/mol / 18.016 g/mol) * 100 = 11%
- For oxygen (O): (16 g/mol / 18.016 g/mol) * 100 = 89%

So, the percent composition of water (H2O) is C. 11% H, 89% O.

I hope this explanation helps! If you have any more questions, feel free to ask.