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3 years ago

How can we calculate empirical formula

Chemistry

1 answer:

Serggg [28]3 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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Would be glad if you helped me!

Sindrei [870]

A) Fe⁰ ----> Fe⁺³ +3e⁻ oxidation | *2
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c) 2Fe⁰ +3Cu⁺² -----> 2Fe⁺³ + 2Cu⁰

6 0

3 years ago

Help me pleaseee. ty if you dooo :))

Lina20 [59]

$\huge \fcolorbox{black}{red}{♛answer♛}$

Hydrogen Fluoride / HF

$\huge\sf\underline{\underline{\red{❥︎ Thanks}}}$

3 0

3 years ago

4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a

yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0

3 years ago

Solid aluminum and gaseous oxygen read in a combination reaction to produce aluminum oxide. 4Al(s) + 3O_2(g) rightarrow 2Al_2O_3

Kryger [21]

Answer:

4.7 g. Option 5 is the right one.

Explanation:

4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)

We convert the mass of reactants to moles, in order to determine the limiting.

2.5 g Al / 26.98 g/mol → 0.092 moles of Al

2.5 g O₂ / 32g/mol → 0.078 moles of O₂

Ratio is 4:3. 4 moles of Al react with 3 moles of O₂

Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂

We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.

Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃

Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles

If we convert the moles to mass, we find the anwer:

0.046 mol . 101.96 g/mol = 4.69 g

5 0

3 years ago

Aluminum reacts with a certain nonmetallic element to form a compound with the general formula AlX. Element X is a diatomic gas

Sergio039 [100]

Answer:

It's nitrogen

Explanation:

cuz it has valence 3 and a diatomic gas at room temperature

5 0

3 years ago

How can we calculate empirical formula​

How can we calculate empirical formula