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2 years ago

How can we calculate empirical formula

Chemistry

1 answer:

Serggg [28]2 years ago

7 0

Explanation:

TO calculate emperical formula , we need percentage composition of the elements and molecular mass of a specific compound which is already given in the question

then create a table with 1.name of element, 2.percentage given and its atomic mass , 3.no. of moles

( %given/at.mass of element) ... then take the least value in the no. of moles you have found now and divide with each no. of moles value .. to get a simple digit number.

You have got the empirical formula

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Question 1 (4 points)

Vilka [71]

To be honest answer might be a

7 0

2 years ago

A .115 L sample of dry air has a pressure of 1.0895633 atm at 377 K. What is the volume of the sample if the temperature is incr

matrenka [14]

Answer:

V= 0.147 L

Explanation:

This is simply the application of combined gas law twice, to find the unknowns.

Combined gas law states that: $PV = nRT$

P= pressure of air

V= volume of air

n= moles of air

R= Universal gas constant ( 0.08205 L atm mol⁻¹ K⁻¹)

T= Absolute temperature in kelvin.

$n=\frac{P * 0.115}{R * 377}$

Now, applying the same gas law at 483K and substituting for n

$V = \frac{P * 0.115}{R * 377} * \frac{R* 483}{P}$

V= $\frac{0.115 * 483}{377}$

V= 0.147 L

4 0

3 years ago

A mixture of hydrocarbons contains 38.3% hexane, C6H14, 13.9% octane, C8H18, and 47.8% decane, C10H22. The mixture is combusted

GaryK [48]

Answer:

52.206 kg

Explanation:

From the given information:

Mass of hexane C6H14 = $19.3*10^3 \ g \times \dfrac{38.3}{100}$

= 7391.9 g

Mass of octane C8H18 = $19.3*10^3 \ g \times \dfrac{13.9}{100}$

= 2682.7 g

Mass of decane C10H22 = $19.3*10^3 \ g \times \dfrac{47.8}{100}$

= 9225.4 g

However, recall that:

number of moles of an atom = mass/molar mass

∴

For hexane, no of moles = 7391.9 g/86.18 g/mol

= 85.77 moles

For octane, no of moles = 2682.7 g/114.23 g/mol

= 23.49 moles

For decane, no of moles = 9225.4 g/142.29 g/mol

= 64.84 moles

Therefore:

number of moles of CO2 produced = (6 × 85.77)+(23.49)+(10×64.84) moles

= 1186.51 moles

Finally, the mass of CO2 produced is:

= 1186.51 mol × 44 g/mol

= 52206.44 g

= 52.206 kg

4 0

2 years ago

The distance between planets varies as they move in their orbits. The minimum distance from the Earth to Mars is about

tigry1 [53]

I would say C 55 000 000 km I’m 80% sure

4 0

3 years ago

6. How many half-filled orbitals are in a bromine atom?<br> 1, 2,3,4

lozanna [386]

Answer:

Bromine has one half filled orbital.

Explanation:

The elements of group 17 are called halogens. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other. As we move down the group in periodic table size of halogens increases that's way fluorine is smaller in size as compared to other halogens elements. Their boiling points also increases down the group which changes their physical states. i.e fluorine is gas while bromine is liquid and iodine is solid.

Electronic configuration of bromine:

₃₅Br = [Ar] 3d¹⁰ 4s² 4p⁵

As it in known that p sub-shell consist of 3 orbitals px, py, pz and each orbital can accommodate only two electrons.

In bromine there are 5 electrons in 4p it means two electrons are present in px two in py ans one in pz. So the half filled orbital is only one.

4 0

3 years ago

How can we calculate empirical formula​

How can we calculate empirical formula