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2 years ago

Calculate the number of molecules present in 3.50 mol feo2. 2.11 x 1024 molecules 2.107 x 1023 molecules 3.50 x 1023 molecules

1 answer:

6 0

Calculate the number of molecules present in 3.50 mol feo2. 2.11 x 1024 molecules 2.107 x 1023 molecules 3.50 x 1023 molecules <u>3.5*(6.02*10^23)=2.107*10^24 molecules</u>

<h3>What is molecules?</h3>

A molecule is the smallest unit of a substance that keeps its content and properties. It is made up of two or more atoms that are joined together by chemical bonds. Chemistry is built on molecules. The element symbol and a subscript indicating the number of atoms are used to identify molecules.

The fundamental building block of an element is an atom. They are made up of an electron-surrounded nucleus. An atom is considered to have valence electrons if its electron shell is not completely complete. A chemical (covalent) connection is created and a lower energy state is entered when two or more atoms join forces to share outer shell valence electrons. In an exothermic reaction, energy is released as atoms bond.

To learn more about molecules from the given link:

brainly.com/question/24722507

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14.2 grams of Na2SO4 is dissolved in water to make a 2.50 L

Trava [24]

Answer:

0.04 M

Explanation:

Given data:

Mass of Na₂SO₄= 14.2 g

Volume of solution = 2.50 L

Molarity of solution = ?

Solution:

Number of moles of Na₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 14.2 g/ 142.04 g/mol

Number of moles = 0.1 mol

Molarity :

Molarity = number of moles of solute / volume of solution in L

Molarity = 0.1 mol / 2.50 L

Molarity = 0.04 M

6 0

3 years ago

EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pK

mihalych1998 [28]

Answer:

Check the explanation

Explanation:

When,

pH = -log[H+] = 3.30

[H+] = $5.0 X 10^{-4} M$

$Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = 7.4 X 10^{-7} ; Ka6 = 4.3 X 10^{-11}$

$alpha[Y^-4] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.56 X 10^{-20} + 3.12 X 10^{-17} + 2 X 10^{-15} + 4 X 10^{-14} + 1.6 X 10^{-13} + 2.34 X 10^{-16} + 2 X 10^{-23}$

= $2.02 X 10^{-13}$

When,

pH = -log[H+] = 10.15

[H+] = $7.08 X 10^{-11} M$

Ka1 = 1 ; Ka2 = 0.0316 ; Ka3 = 0.01 ; Ka4 = 0.002 ; Ka5 = $7.4 X 10^{-7}$ ; Ka6 = $4.3 X 10^-11$

$alpha[Y^{-4}] = [H+]^6 + Ka1[H+]^5 + Ka1Ka2[H+]^4 + Ka1Ka2Ka3[H+]^3 + Ka1Ka2Ka3Ka4[H+]^2 + Ka1Ka2Ka3Ka4Ka5[H+] + Ka1Ka2Ka3Ka4Ka5Ka6$

= $1.26 X 10^{-61} + 1.8 X 10^{-51} + 8.1 X 10^{-43} + 1.12 X 10^{-34} + 3.17 X 10^{-27} + 3.3 X 10^{-23} + 1.83 X 10^{-23}$

= $5.12 X 10^{-23}$

4 0

3 years ago

A pharmaceutical company is making a large volume of nitrous oxide (NO). They predict they will be able to make a maximum amount

iragen [17]

Answer:

The answer is "Option B"

Explanation:

From the query, the following knowledge is derived:

Yield in percentage = 47%

Performance of theory = 4860 g

Actual yield Rate =?

The percentage return is defined simply by the ratio between both the real return as well as the conceptual return multiplied by the 100. It's also represented as numerically:

$Rate = \frac{Existing \ Rate} {Theoretical \ Rate} \times 100$