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Liula [17]
4 years ago
5

A compound contains 38.7% K, 13.9% N, and 47.4% O by mass. What is the empirical formula of the compound?

Chemistry
1 answer:
inysia [295]4 years ago
8 0
K ---> 38.7 g / 39.1 g/mol = 0.99
N ---> 13.9 g / 14.0 g/mol = 0.99
O ---> 47.4 g / 16.0 g/mol = 2.96

Divide by smallest:

K ---> 0.99 / 0.99 = 1
N ---> 0.99 / 0.99 = 1
O ---> 2.96 / 0.99 = 3

KNO3
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We are given with the mass of Arsine (AsH_{3}

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moles=\frac{15}{77.92}=0.192mol


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What impact would a sudden increase in the price of wood have on producers and consumer? How might each group respond to the cha
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6 0
3 years ago
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What is the difference between a heterogeneous and homogeneous mixture? A. A heterogeneous can only be solid, and a homogeneous
muminat

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3 0
4 years ago
Please help!!!!! I need the correct answer quickly!!!
xxMikexx [17]

Answer :

The oxidation state of oxygen (O) in OF_2  is, (+2)

The oxidation state of carbon (C) in CO  is, (+2)

The oxidation state of nitrogen (N) in K_3N  is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(a) The given compound is, OF_2

Let the oxidation state of 'O' be, 'x'

x+2(-1)=0\\\\x-2=0\\\\x=+2

The oxidation state of oxygen (O) in OF_2  is, (+2)

(b) The given compound is, CO

Let the oxidation state of 'C' be, 'x'

x+(-2)=0\\\\x-2=0\\\\x=+2

The oxidation state of carbon (C) in CO  is, (+2)

(c) The given compound is, K_3N

Let the oxidation state of 'N' be, 'x'

3(+1)+x=0\\\\3+x=0\\\\x=-3

The oxidation state of nitrogen (N) in K_3N  is, (-3)

7 0
3 years ago
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