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2 years ago

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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Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.

o-na [289]

Answer:

pKa = 7.46

Explanation:

1) Data:

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

2) Strategy:

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

3) Solution:

a) pH

pH = - log [H₃O⁺]

4.64 = - log [H₃O⁺]

$[H_3O^+]= 10^{-4.64} = 2.29.10^{-5}$

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0

2 years ago

The decomposition of nitramide, O2NNH2, in water has the chemical equation and rate law O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H

FinnZ [79.3K]

Answer:

It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law

Explanation:

Please, observe the solution in the attached Word document.

Download docx

7 0

2 years ago

Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t

mars1129 [50]

Answer:

Explanation:

N₂ + 3H₂ = 2 NH₃

1 vol 2 vol

786 liters 1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

$\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }$

$V_2 =303.44$ liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0

2 years ago

1. If a car is traveling 90 mi/hr, how many feet will the car travel in 1 sec? In 5 sec?

Fudgin [204]

Answer:

The car travel 660 feet

Explanation:

First convert the Speed into Feet/sec

Speed of the car = 90 mi/hr

1 mi = 5280 ft

In 90 miles = 90 x 5280 = 475,200 feet

1 hr = 60 min

1 min = 60 sec

So ,

1 hr = 60 x 60 sec = 3600 sec

The speed is defined as the distance traveled by the object in unit time. The formula of speed is :

$Speed =\frac{distance}{time}$

$Speed =\frac{90mi}{1hr}$ ... given

$Speed =\frac{475200 ft}{3600sec}$

Speed = 132 ft/sec

Now, time = 5 sec

$Speed =\frac{distance}{time}$

$distance =speed\times time$

$distance = 132ft/sec\times 5sec$

Distance = 660 feet

3 0

2 years ago

C6H12O6+6O2→6CO2+6H2O

babunello [35]

Answer:

The answer that is true is A

Explanation:

The first thing to do is to perform stoichiometry relationships and calculations.

First relationship: 0.3 moles of CO₂ are produced

6 mol CO₂ _____ 1 mol C₆H₁₂O₆

0.3 mol CO₂ _____ X = 0.05 mol of C₆H₁₂O₆

Calculation: 0.3 mol x 1 mol / 6 mol = 0.05 mol of C₆H₁₂O₆

Second relationship: 0.60 moles of CO₂ are produced

6 mol CO₂ _____ 1 mol C₆H₁₂O₆

0.6 mol CO₂_____ X = 0.1 mol of C₆H₁₂O₆ 

Calculation: 0.6 mol x 1 mol / 6 mol = 0.1 mol of C₆H₁₂O₆

6 mol CO₂ _____ 6 mol O₂

0.6 mol CO₂ _____ X = 0.6 mol of O₂

Calculation: 0.6 mol x 6 mol / 6 mol = 0.6 mol of O₂

Then, the answer that is true is A, because in order for 0.6 moles of CO₂ to be produced, there must initially be 0.1 mol of C6H12O6 as calculated.

The B is not correct because for more CO₂ to be produced, more C₆H₁₂O₆ must react, this is defined according to Proust's law or law of defined proportions, stated as: “When the combination of two or more elements occurs to form a new compound, they always do it in a mass relationship that is completely constant”. It is by this law that no more CO₂ can be produced with the same amount of C₆H₁₂O₆ reagent.

The C is not correct either, although the temperature and volume of the container are the same, the O₂ is in excess to react, so 0.6 mol of it is needed to produce 0.6 mol of CO₂, according to the calculations made .

The D is not correct because it would also not respect the law of defined proportions or Proust's law. Although O₂ is in excess, it cannot react more than the stoichiometric ratio. Simply, if it is in excess, the rest of the O₂ will remain unreacted.

6 0

2 years ago