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vagabundo [1.1K]
3 years ago
14

In a reaction involving Iron Fe and Oxygen It was determined that 4.166g of Iron reacted with 1.803 g of Oxygen. Determine the e

mpirical formula of the compound that resulted
Chemistry
1 answer:
ziro4ka [17]3 years ago
7 0

Answer:

Empirical formula is Fe₂O₃.

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number.

Given data:

Mass of iron = 4.166 g

Mass of oxygen = 1.803 g

Empirical formula = ?

Solution:

Number of gram atoms of Fe = 4.166 / 55.845 = 0.075

Number of gram atoms of O = 1.803 / 15.999 = 0.113

Atomic ratio:

                Fe                        :             O

              0.075/0.075          :           0.113/0.075

                   1                          :               1.5

Fe :  O = 1 : 1.5

Fe :  O = 2(1 : 1.5)

Empirical formula is Fe₂O₃.

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Electrons are moving in energy levels around nucleus.

The electron has a mass that is approximately 1/1836 that of the proton.

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What are two types of matter are pure substances
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4 0
3 years ago
An element has an atomic number of 18 and an atomic mass of 40. The number of neutrons in the nucleus of an atom of this element
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7 0
3 years ago
What mass of sucrose (C12H22O11) should be combined with 546 g of water to make a solution with an osmotic pressure of 8.80 atm
lesya [120]

<u>Answer:</u> The mass of sucrose required is 69.08 g

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 8.80 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (sucrose) = ?

Molar mass of sucrose = 342.3 g/mol

Volume of solution = 564 mL    (Density of water = 1 g/mL)

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = Temperature of the solution = 290 K

Putting values in above equation, we get:

8.80atm=1\times \frac{\text{Mass of sucrose}\times 1000}{342.3\times 546}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 290K\\\\\text{Mass of sucrose}=\frac{8.80\times 342.3\times 546}{1\times 1000\times 0.0821\times 290}=69.08g

Hence, the mass of sucrose required is 69.08 g

5 0
3 years ago
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