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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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Micky Mo is a 64-year-old male admitted to the emergency room for asthma. His laboratory results are as follows: pH 7.31, pCO2 h

Salsk061 [2.6K]

Answer:

Micky Mo is suffering from respiratory acidosis.

Explanation:

The pCO2 level in micky"s body is higher than normal it means the excess amount of CO2 will reacts with water to generate carbonic acid(H2CO3).

On the other hand according to the question total HCO3- also higher than normal.As a result the excess HCO3- will react with proton to form carbonic acid which is in turn dissociate to generate CO2 and H2O to maintain normal acid base homeostasis.

From that point of view it can be said Micky Mo is suffering from respiratory acidosis.

5 0

1 year ago

4.8g of calcium is added to 3.6g of water. The following reaction occurs

notka56 [123]

Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present

q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present

sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present

8 0

1 year ago

A 3.5 M KNO3 solution contains 3.5 miles of KNO3 dissolved in A) 1.0 liter of solution B) 1.0 liter of solvent C) 3.5 liters of

Vladimir [108]

N=3.5 mol
c=3.5 mol/L

n=cv

v=n/c

v=3.5/3.5=1.0 L

A) 1.0 liter of solution

4 0

1 year ago

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

Keith_Richards [23]

Answer:

1) 0.009 61 g C; 2) 0.008 00 mol C

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r: 12.01 44.01

C + ½O₂ ⟶ CO₂

m/g: 0.352

1) <em>Mass of C </em>

Convert grams of CO₂ to grams of C

44.01 g CO₂ = 12.01 g C

Mass of C = 0.352 g CO₂ × 12.01 g C/44.01 g CO₂

Mass of C = 0.009 61 g C

2) <em>Moles of C </em>

Convert mass of C to moles of C.

1 mol C = 12.01 g C

Moles of C = 0.00961 g C × (1 mol C/12.01 g C)

Moles of C = 0.008 00 mol C

All the carbon comes from Compound A, so there are 0.008 00 mol C in Compound A.

7 0

2 years ago

For the reaction, 2cr2+ + cl2(g) ---> 2cr3+ + 2cl- e cell (standard conditions) = 1.78v calculate ecell (standard conditions)

PolarNik [594]

Answer:

-1.78 V

Explanation:

There are several rules required to calculate the cell potential:

given standard cell potential, we may reverse the equation: the products of a given reaction become our reactants, while reactants become our products in the reversed equation. For a reversed equation, we change the sign of the cell potential to the opposite sign;
if we multiply the whole equation by some number, this doesn't influence the cell potential value. It only produces a different expression in the equilibrium constant.

That said, notice that the initial reaction with respect to the final reaction is:

reversed: chromium(III) cation and chloride anion become our reactants as opposed to the products in the initial reaction, so we change the sign of the cell potential to a negative value of -1.78 V;
each coefficient is multiplied by a fraction of $\frac{1}{2}$ . It doesn't influence the value of the cell potential.

Thus, we have a cell of E = -1.78 V.

3 0

2 years ago