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2 years ago

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

Chemistry

1 answer:

REY [17]2 years ago

5 0

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

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What do you expect will occur as air masses move over cold locations? Hot locations? How does this movement of air masses affect

Maksim231197 [3]

Answer:

Explanation:

When air masses will move over cold location then bottom layer of air cools and become more dense. Due to its high density it is trapped near the ground then it flow towards the equator.

When it moves over hot location then bottom layer gets hot and lighter. Then it moves towards poles.

It changes the temperature and humidity of the climate. making hoter region coll and cooler region a bit hot.

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3 years ago

When 8.00x10^22 molecules of ammonia react with 7.00x10^22 molecules of oxygen according to the chemical equation shown below, h

Pavel [41]

NH₃:

N = 8*10²²
NA = 6.02*10²³

n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol

O₂:

N=7*10²²
NA = 6.02*10²³

n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol

4NH₃ + 3O₂ ⇒ 2N₂ + 6H₂O
4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant

N₂:

n = 0.0665mol
M = 28g/mol

m = n*M = 0.0665mol*28g/mol = 1,862g

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3 years ago

A dilute solution of bromine in carbon tetrachloride behaves as an ideal-dilute solution. The vapour pressure of pure CCl4 is 33

ANEK [815]

Explanation:

The given data is as follows.

Vapour pressure of pure $CCl_{4}$ = 33.85 Torr

Temperature = 298 K

Mole fraction of $Br_{2}$ = 122.36 torr

Therefore, calculate the vapor pressure of $Br_{2}$ as follows.

Vapour pressure of $Br_{2}$ = mole fraction of $Br_{2}$ x K of $Br_{2}$

= 0.050 x 122.36 Torr

= 6.118 Torr

So, vapor pressure of $Br_{2}$ is 6.118 Torr .

Now, calculate the vapor pressure of carbon tetrachloride as follows.

Vapour pressure of $CCl_{4}$ = mole fraction of $CCl_{4}$ x Pressure of $CCl_{4}$

= (1 - 0.050) × 33.85 Torr

= 32.1575 Torr

So, vapor pressure of $CCl_{4}$ is 32.1575 Torr .

Hence, the total pressure will be as follows.

= 6.118 Torr + 32.1575 Torr

= 38.2755 Torr

Therefore, composition of $CCl_{4}$ = $\frac{32.1575 Torr}{38.2755 Torr}$

= 0.8405

Composition of $CCl_{4}$ is 0.8405 .

And, composition of $Br_{2}$ = $\frac{6.118 Torr}{38.2755 Torr}$

= 0.1598

Composition of $Br_{2}$ is 0.1598 .

6 0

3 years ago

60 POINTS + BRAINLIEST !! PLEASE HELP!!

uranmaximum [27]

To get moles. divide mass by molar mass.Molar mass of
Na is 23
and for Cl is 35.5.
the total molar mass of NaCl is 23+35.5 = 58.5mol/gUse the mass and divide by this number30.22g divide by 58.5mol/g and you will get 0.5166 mole.
Since the molecule has 1 Na to 1 Cl, and that the number of moles for NaCL is 0.5166. All of them would be 0.5166molesNa = 0.5166 x 1 = 0.5166molesCl = 0.5166 x 1 = 0.5166moles
to get number of atoms. Multiply your mole by Avogadro number which is 6.022x10^23Na = 0.5166 x 6.022E23 = 3.111x10^23Cl = 0.5166 x 6.022E23 = 3.111x10^23

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