Which part of the electromagnetic spectrum is nearest to x rays

A nitrogen oxide, containing 53.85% N, acts as a vasodilator, lowering blood pressure in the human body. What is its empirical f

jenyasd209 [6]

The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is $N_4O_3$ . Empirical formula is the smallest whole number ratios of elements in a compound.

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1. Get the mass percent data for all the elements that make up the compound.

2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3. Convert the mass of each element to number of moles.

4. Divide each calculated mole by the smallest mole value.

5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

$moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89$

The moles may be written as the temporary subscripts of the elements in the compound as follows:

$N_{3.85}O_{2.89}$

STEP 4: Divide the moles by the smallest mole value.

$N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\$

STEP 5: Multiply the subscripts by a number that would give the smallest whole number ratio.

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

$N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}$

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

$mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}$

$mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%$

Since the values are the same from the given, the answer is correct.

LEARN MORE

Writing Chemical Formula brainly.com/question/4697698
Naming Compounds brainly.com/question/8968140
Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, compounds

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Lighting a match changes __________ and _____________ energy into heat.

Assoli18 [71]

Lighting a match changes chemical energy into heat.

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The hydrolysis of ethyl acetate in alkaline conditions yields acetate ion and ethanol. The rate of the reaction increases 6.37 t

sweet [91]

Answer:

Explanation:

Hydrolysis is a chemical decomposition involving breaking of a bond and the addition of elements

of water. In this hydrolysis of ester (ethyl acetate) with an alkali (sodium hydroxide), HCl was

used as catalyst to accelerate it. 1ml and 2ml of ethyl acetate was injected separately into the

500ml reactor vessels which contains 0.05N of NaOH and thoroughly mixed. At regular time

interval, 0, 10, 20, 30, 40, and 90 minutes, 25ml of each of the samples were withdrawn into a

250ml conical flask containing 0.05N HCl, and titrated against 0.05N NaOH solution using

phenolphthalein as indicator. From the titre values, the hydrolysis involving the 1ml ethyl acetate

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A package within package can be best defined as which of the following

Citrus2011 [14]

Answer:

double-protected

Explanation:

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Match the theoretical motions of particles with their associated phases. Solid, Gas, Liquid. 1. oscillation about a fixed point

viktelen [127]

The correct combination of theoretical motions of particles with their associated phases is given below:

1) Solid : oscillation about a fixed point . Because solids do vibrate (atoms/molecules of a solid oscillate around their mean positions)
2) Gas : translation, vibration, rotation. The gas molecules are free to rotate, oscillate & vibrate. 
3) Liquid: rotation and vibration. The liquid molecules are free to rotate & vibrate.

A more clear pictures (at atomic & molecular level is shown below):

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2 years ago

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