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3 years ago

What information is needed to calculate the percent composition of a compound

2 answers:

6 0

Percent Composition by Mass. Percent composition is calculated from a molecular formula by dividing the mass of a single element in one mole of a compound by the mass of one mole of the entire compound. This value is presented as a percentage.

The best answer. :)

marysya [2.9K]3 years ago

5 0

Answer:

Molecular formula, molecular mass and the mass of each element in the molecule.

Explanation:

Percent Composition by Mass is the percentile of each element present in a molecule.

It is calculated by dividing the molar mass of a element by the molecule molar mass, multiplied by 100

Eg the mass percent compositionof carbon in carbon dioxide is 27.27%

Molecular formula: CO2

Molecular Mass: 44g/mol

C molar mass: 12g/mol

O molar mass: 16g/mol

$Percent_{carbon} =\frac{12}{44}*100=27.27$

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If p = mv, what does v =? (Solve for v)

Darina [25.2K]

if p=mv, then v= p/m

6 0

4 years ago

a radioisotope has a half life of three hours how much of a 120g sample remains after 9 hours. what will the g amount be after 3

aksik [14]

$\huge\boxed{♔︎Answer♔︎}$

60g after 3 hours, 30g after 6 hours and 15g after 9 hours

Explanation:

Weight of the radioactive sample = 120g

half life time period = 3 hours

(a) The weight of sample after 3 hours

$\textsf{ No. of half lives} = \sf \cancel\frac{3}{3} = 1$

The fraction of sample left

$\sf { \frac{1}{2} }^{1} = \frac{1}{2}$

Mass of the sample left

$\sf \frac{1}{2} \times 120 = \cancel\frac{120}{2} = 60g$

60g of sample is left after 3 hours

(b) The weight of sample after 6 hours

$\textsf{ No. of half lives} = \sf \cancel\frac{6}{3} = 2$

The fraction of the sample left

$\sf { \frac{1}{2} }^{2} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Mass of the sample left

$\sf \frac{1}{4} \times 120 = \cancel\frac{120}{4} = 30g$

30g of sample is left after 6 hours

$\textsf {No. of half lives} = \sf \cancel\frac{9}{3} = 3$

The fraction of sample left

$\sf { \frac{1}{2} }^{3} = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

Mass of sample left

$\sf \frac{1}{8} \times 120 = \cancel \frac{120}{8} = 15g$

15g of sample is left after 9 hours.

3 0

2 years ago

When a sample of aqueous hydrochloric acid was neutralized with aqueous sodium hydroxide in a calorimeter, the temperature of 10

Annette [7]

6.50x10^3 calories. Now we have 4 pieces of data and want a single result. The data is: Mass: 100.0 g Starting temperature: 25.0Â°C Ending temperature: 31.5Â°C Specific heat: 1.00 cal/(g*Â°C) And we want a result with the unit "cal". Now you need to figure out what set of math operations will give you the desired result. Turns out this is quite simple. First, you need to remember that you can only add or subtract things that have the same units. You may multiply or divide data items with different units and the units can combine or cancel each other. So let's solve this: Let's start with specific heat with the unit "cal/(g*Â°C)". The cal is what we want, but we'ld like to get rid of the "/(g*Â°C)" part. So let's multiply by the mass: 1.00 cal/(g*Â°C) * 100.0 g = 100.0 cal/Â°C We now have a simpler unit of "cal/Â°C", so we're getting closer. Just need to cancel out the "/Â°C" part, which we can do with a multiplication. But we have 2 pieces of data using "Â°C". We can't multiply both of them, that would give us "cal*Â°C" which we don't want. But we need to use both pieces. And since we're interested in the temperature change, let's subtract them. So 31.5Â°C - 25.0Â°C = 6.5Â°C So we have a 6.5Â°C change in temperature. Now let's multiply: 6.5Â°C * 100.0 cal/Â°C = 6500.0 cal Since we only have 3 significant digits in our least precise piece of data, we need to round the result to 3 significant figures. 6500 only has 2 significant digits, and 6500. has 4. But we can use scientific notation to express the result as 6.50x10^3 which has the desired 3 digits of significance. So the result is 6.50x10^3 calories. Just remember to pay attention to the units in the data you have. They will pretty much tell you exactly what to add, subtract, multiply, or divide.

4 0

3 years ago

Is Salt and pepper an example of a homogenous mixture?

alisha [4.7K]

Answer:

no it is a heterogeneous mixture because salt and pepper are not mixed uniformly.

4 0

4 years ago

Which statement is incorrect regarding the reaction of benzene with an electrophile? View Available Hint(s) A) The carbocation i

Leno4ka [110]

Answer:

The carbocation intermediate reacts with a nucleophile to form the addition product.

Explanation:

The reaction of benzene with an electrophile is an electrophillic substitution reaction. Here the electrophile replaces hydrogen. There is no formation of carbocation as intermediate in the reaction. Infact there is transition state where the electorphile attacks on benzene ring and at the same time the hydrogen gets removed from the benzene. So a transition carbocation is formed.

The general mechanism is shown in the figure.

i) Attack of the electrophile on the benzene (which is the nucleophile)

ii) The carbocation intermediate loses a proton from the carbon bonded to the electrophile.

iii) the carbocation formation is the rate determining step.

iv) There is no formation of addition product.

Thus the wrong statement is

The carbocation intermediate reacts with a nucleophile to form the addition product.

3 0

4 years ago