What is the molar mass of C3H8O2? (Round to three significant figures)

Answers

Answer 1

Answer:

The molar mass of [tex]C_3H_8O_2[/tex] (rounded to three significant figures) is 76.1 amu.

Step-by-step explanation:

Hey there!

To answer this question, you will need access to a periodic table. You need to know the atomic masses of carbon (C), hydrogen (H), and oxygen (O) in order to answer this question. With that out of the way, let's begin!

We have a chemical formula: [tex]C_3H_8O_2[/tex].

The letters are element symbols, which are basically the code for an element. Think of it as an abbreviation or something similar to an ID.The subscripts (numbers) help us identify how many atoms of each element are included in the substance.

The first thing we should do is locate C, H, and O on a periodic table and figure out the atomic masses for each element. This will help us determine the total molar mass of the substance.

Determining the Atomic Mass of C

If we locate C on a periodic table, we see that its atomic mass is roughly 12.01 amu (atomic mass units). Then, we have a subscript of 3, so we simply multiply the subscript by the given atomic mass in order to find the atomic mass of the three atoms of carbon.

[tex]3 \times 12.01 = 36.03 \text{ amu}[/tex]

This means that C accounts for 36.03 amu of the molar mass of the substance.

Determining the Atomic Mass of H

Using the same principle as before, we want to find the atomic mass of one atom of H. This is 1.008, so we will multiply 1.008 by the eight atoms (pay attention to the subscript) in order to find the atomic mass of hydrogen in the substance.

[tex]8 \times 1.008 = 8.064 \text{ amu}[/tex]

This means that H accounts for 8.064 amu of the molar mass of the substance.

* An important note to add is that we should never round our values until the end of the problem so we make sure to get as close as possible (this is why you learn about accuracy and precision in a chemistry class).

Determining the Atomic Mass of O

Finally, we want to find the atomic mass of O in the substance, so we'll use the same method as before and locate the atomic mass of one atom of oxygen. We find that this is 15.999, which would definitely round to 16.00, but remember the rules for rounding in chemistry in order to be as accurate as possible. Since our subscript is 2, we know that there are two oxygen atoms, so we will multiply the atomic mass of oxygen by 2.

[tex]2 \times 15.999 = 31.998 \text{ amu}[/tex]

This means that O accounts for 31.998 amu of the molar mass of the substance.

Summation of the Determined Atomic Masses

After you have determined the atomic masses of the atoms in the chemical substance, you have to add these together. This is performed with simple addition.

C = 36.03 amuH = 8.064 amuO = 31.998 amu

Now, let's add these together.

[tex]36.03 \text{ amu} + 8.064 \text{ amu} + 31.998 \text{ amu} = 76.092 \text{ amu}[/tex]

Therefore, the total atomic mass is 76.092 amu.

Rounding to Meet Significant Figures Requirements

Significant figures are used to determine the accuracy of a problem. A basic rule is to use scientific notation to determine how to meet the significant figure request. Since we are rounding to three significant figures, we want to make sure these three figures are significant. There are three basic rules to remember when computing significant figures:

Any zeroes to the left of the first non-zero number (ex: 0.0981) are insignificant.Any zeroes "trapped" in between other non-zero numbers (ex: 9.81075) are significant.Any zeroes to the right of the last non-zero number (ex: 5900) are significant ONLY IF they are added after a decimal point and included in the measurement (ex: 5900.)

Therefore, since we need three significant figures, we have to round our value first and then determine the appropriate number of significant figures.

76.092 will round to 76.09 first, and then round to 76.1. We cannot use 76.09 since there is a trapped zero, which is considered significant.We must use 76.1 since there are three significant figures in the value.

Final Answer

Since we have determined that 76.1 fits our requirements for three significant figures, we can safely determine that the molar mass of [tex]C_3H_8O_2[/tex], as determined, is 76.1 amu.

Answer 2
Mass of 1 mole of a substance is called its molar mass.The molar mass of a molecular substance is equal to the molecular mass of the substance expressed in the units of g/mol.Atomic mass of Carbon (C) = 12.01 uAtomic mass of Hydrogen (H) = 1.008 uAtomic mass of Oxygen (O) = 15.999Therefore, molar mass of C3H802

= 3 × Mass of C + 8 × Mass of H + 2 × Mass of O

= (3 × 12.01 + 8 × 1.008 + 2 × 15.999) g/mol

= (36.03 + 8.064 + 31.998) g/mol

= 76.092 g/mol

Answer:

76.092 g/mol

Hope you could get an idea from here.

Doubt clarification - use comment section.


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Answer:

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Answers

Answer:

Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2

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Answers

Answer:

The photon is directly proportional to frquency

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h is plancks constant and

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A

Explanation:

Edge 2022

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Answer:

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Answer:

Because to them it was just a little bit of fiction. so they don't really think it's necessary to do experiments or make detailed observation

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In comparing groups, the first or the second, which group was more precise or was the precision the same

Answers

Comparing the data obtained online, the first group is more precise

The reason the above selection is correct is as follows;

Question: Parts of the question appear missing from a similar question online and included here;

First Group:

[tex]\begin{array}{|c|c|c|c|c|c|c|} Team \ a &Team \ b&Team \ c&Team \ d&Team \ e&Team \ f&Team \ g\\2.77 \ cm&2.60 \ cm&2.80 \ cm&2.65 \ cm&2.75 \ cm&2.65 \ cm& 2.68 \ cm\end{array}\right][/tex]

Second Group

[tex]\begin{array}{|c|c|c|c|c|c|c|} Team \ a &Team \ b&Team \ c&Team \ d&Team \ e&Team \ f&Team \ g\\2.70 \ cm&2.78 \ cm&2.62 \ cm&2.65 \ cm&2.75 \ cm&2.80 \ cm& 2.60 \ cm\end{array}\right][/tex]

Calculations for the first group:

The average = (2.77 + 2.60 + 2.80 + 2.65 + 2.75 + 2.65 + 2.68)/7 = 2.7

The range = 2.80 - 2.60 = 0.20

The approximate ± range of the average = ±0.2/2 = ±0.1

The precision of the first measurement is 2.7 ± 0.1 cm

Calculations for the second group:

The average = (2.70 + 2.78 + 2.62 + 2.65 + 2.75 + 2.80 + 2.60)/7 = 2.7

The range = 2.80 - 2.60 = 0.20

The approximate ± range from the average = ±0.2/2 = ±0.1

The range of values from the average is approximately ±0.1 cm

The precision of the first measurement is 2.7 ± 0.1 cm

Method to determine precision:

Precision is given by the finding the average deviation and the standard deviation as follows;

[tex]\mathbf{Average \ deviation} = \dfrac{\sum \left | x - \mu \right |}{n}[/tex]

[tex]\mathbf{Standard\ deviation} =\sqrt{\dfrac{\sum \left ( x - \mu \right )^2}{n}}[/tex]

Where;

μ = The mean or average = 2.7

n = The number of items (count) of the data = 7

For the first group, we have;

[tex]\mathbf{\dfrac{\sum \left | x - \mu \right |}{n}} = \dfrac{0.07 + 0.1 + 0.1 + 0.05 + 0.05 + 0.05 + 0.02 }{7} = \mathbf{0.062857}[/tex]

The average deviation of the first group = 0.062857

∑(x - μ)² ≈ 0.0328

[tex]\mathbf{The \ standard\ deviation} =\sqrt{\dfrac{0.0328}{7}} \approx \mathbf{0.068452}[/tex]

For the second group, we have;

[tex]\mathbf{\dfrac{\sum \left | x - \mu \right |}{n}} = \dfrac{0.0+ 0.08 + 0.08 + 0.05 + 0.05 + 0.1 + 0.1 }{7} = \mathbf{0.06571428571}[/tex]

∑(x - μ)² ≈ 0.0378

[tex]\mathbf{The \ standard\ deviation} =\sqrt{\dfrac{0.0378}{7}} \approx \mathbf{0.073485}[/tex]

By using the average deviation, and standard deviation values, the deviation of the second group is more than the first group and therefore the first group is more precise

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How is the atomic mass of an element determined?

A. by adding the masses of the protons and electrons

B. by adding the masses of the neutrons and electrons

C. by determining the number of protons

D. by adding the masses of the protons, neutrons, and electrons

Answers

Answer:

D

Explanation:

Atomic mass is the total mass of protons, neutrons and electrons of an atom at rest

Calculate the no. of moles in 25gm of calcium phosphate

Plz Fast Urgent

Answers

Answer:

0.162 mol

Explanation:

n=m/M

M= 20*3 + 15*2 + 8*4*2

M=154 g/mol

n=25g/154g/mol

n=0.162 mol

The number of moles in 25 grams of calcium phosphate is 0.162 mol.

What are moles?

Mole is a SI unit that is used to measure the quantity of any substance in a specific amount. To calculate the moles, we will divide the mass of the substance by the molar mass of that substance.

n = m/M

Calcium phosphate is a chemical compound that contains calcium and phosphorus or phosphoric acid, and it is used to treat the low calcium in the body and thyroid and parathyroid.

The molar mass of calcium sulfate is 154 g/mol

The mass of calcium sulfate is 25 grams

Putting the value in the equation

n = m/M

n = 25g / 154 g/mol = 0.162 mol

Thus, the moles of calcium sulfate are 0.162 mol.

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What would the car weigh on Earth with a gravitational pull of 9.8 N/kg?

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Answer:

Explanation:s the gravitational field strength on Earth's 9.8 N kg?

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Yes. Simple explanation, air is real, but you can’t see it. You can breathe it (sometimes smell it), feel it, etc. Sorry if this answer doesn’t help much.

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Answer:

the surface area of a solid reactant.

temperature.  

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The strength of van der Waals forces increases as:
A. molecular size decreases
B. molecular size increases
C. number of electrons increases
D. number of electrons decreases

Answers

Answer:

B and C

Explanation:

is it multiple choice question??

Van der Waals forces are the interaction between the molecules and the atoms. It increases with an increase in the molecular size and the number of electrons. Thus, options B and C are correct.

What are Van der Waals force?

Van der Waals forces are the weak forces of attraction and repulsion present between the atoms and the molecules of the compounds. They are also called London Dispersion Forces and dipole-dipole forces.

An increase in the number of electrons and size of the molecule increases the surface area and the electron cloud which in turn, increases the Van der Waals force.

The intermolecular force increases as in large atoms or molecules the valence are held loosely as they are far from the nucleus or the center of the atom.

Therefore, the size and the electron increase the Van der Waals force.

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