Which of the following will have the greatest proportion by mass of nitrogen?

musickatia [10]

Answer:

Natural science can be divided into two main branches

Explanation:

life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.

8 0

3 years ago

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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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A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg

Dafna1 [17]

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

8 0

3 years ago

If a compound has a composition of 82% nitrogen and 18% hydrogen, what is the empirical formula for this compound

Damm [24]

Answer: The empirical formula for the given compound is $NH_3$

Explanation : Given,

Percentage of H = 18 %

Percentage of N = 82 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of H = 18 g

Mass of N = 82 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{18g}{1g/mole}=18moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{82g}{14g/mole}=5.8moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.

For Hydrogen = $\frac{18}{5.8}=3.10\approx 3$

For Nitrogen = $\frac{5.8}{5.8}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of H : N = 3 : 1

Hence, the empirical formula for the given compound is $NH_3$

3 0

3 years ago

In which type of climate would the soil most likely be well developed and have lots of organic matter?

sleet_krkn [62]

I suppose it would be forest because in order to have organic matter the soil needs to be rich and fertile,therefore it is forest.

3 0

3 years ago