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nika2105 [10]
2 years ago
14

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

Chemistry
1 answer:
mojhsa [17]2 years ago
8 0

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

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Answer:

The Michaelis‑Menten equation is given as

v₀ = Kcat X [E₀] X [S] / (Km + [S])

where,

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3 0
2 years ago
A vessel of capacity 400 cc filled with chlorine under 80 cm is connected by a narrow tube and stopcock with another vessel of c
guapka [62]

Answer:

The pressure when the stopcock is opened is opened is 87.783 cm

Explanation:

The given parameters of the question are;

The volume of the vessel of chlorine = 400 cc

The pressure of the vessel of chlorine = 80 cm

The volume of the vessel of chlorine = 250 cc

The pressure of the vessel of chlorine = 100 cm

Daltons law of Partial Pressures states that the total pressure exerted by a volume of a mixture of gases is equal to the partial pressures exerted by the individual gases in the mixture with respect to the given volume

Therefore;

The total volume of the mixture = 400 cc + 250 cc = 650 cc

The partial pressure exerted by the chlorine gas in the total volume is given by Boyles law as follows;

P₁·V₁ = P₂·V₂

P₂ = P₁·V₁/V₂

Where;

P₁ = 80 cm = The pressure in volume V₁ = 400 cc

P₂ₓ = The partial pressure of chlorine in volume V₂ = 650 cc

Substituting, we have;

P₂ₓ = 80 × 400/650 ≈ 49.321 cm

Similarly, the partial pressure exerted by the nitrogen gas in the total volume is given by Boyles Law as follows;

P₂ₐ = P₁·V₁/V₂

Where;

P₁ = 100 cm = The pressure in volume V₁ = 250 cc

P₂ₐ = The partial pressure of nitrogen in volume V₂ = 650 cc

Substituting, we have;

P₂ₐ = 100× 250/650 ≈ 38.462 cm

The pressure of the combined gas, P, when the stopcock is opened is opened is given by Dalstons Law of partial pressure as P = P₂ₐ + P₂ₓ

Therefore, the pressure, P when the stopcock is opened is opened = 49.321 cm + 38.462 cm = 87.783 cm

3 0
1 year ago
Plseas help Calculate the amount of moles in 57.6 Liters of Carbon Dioxide.
slavikrds [6]
Convert 57.6 L to dm3 and divide it by 24
8 0
2 years ago
Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50 Ti (5.3%). What is the average atomic
zhannawk [14.2K]
(46x8.0)+(47x7.8)+(48x73.4)+(49x5.5)+(50x5.3) = 4792.3

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6 0
2 years ago
Time Remaining: 1:27:31 If 50.0 g of H₂ and 100.0 g of O₂ react, how many moles of H₂O can be produced in the reaction below? 2
Novay_Z [31]

Answer:

Explanation:

2 H₂(g) + O₂(g) → 2 H₂O(g

2 moles     1 mole      2 mole

50 g of H₂ = 50 /2 = 25 moles of H₂

100 g of O₂ = 100 / 32 = 3.125 moles of O₂

So oxygen is the limiting reagent .

3.125 moles of O₂ will react with 6.25 moles of H₂ to give 6.25 moles of H₂O .

Hence moles of H₂O produced = 6.25 moles .

6 0
2 years ago
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