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Luba_88 [7]
2 years ago
8

What is the percent composition of each element within the compound​

Chemistry
2 answers:
kogti [31]2 years ago
6 0

Answer:

The percentage composition of a given compound is defined as <u>the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. </u>Here, the quantity is measured in terms of grams of the elements present.

please give me brainliest

iren2701 [21]2 years ago
6 0

The ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100

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stiks02 [169]

Answer:

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5 0
3 years ago
Read 2 more answers
What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O
Ipatiy [6.2K]

Answer:

Mass = 13.23 g  

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂     →      3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

              O₂           :          C₃H₈

               5            :            1

             1.5            :          1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g  

6 0
3 years ago
Please help with everything
Vanyuwa [196]
Search the question up on google and it will show up.
5 0
3 years ago
GUYS I NEED THIS ASAP ​
liberstina [14]
It’s DEFINITELY 2 like DEFINITELY
8 0
3 years ago
15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

7 0
3 years ago
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