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Drupady [299]
3 years ago
11

Calculate the percent by mass of each element in

Chemistry
1 answer:
Zepler [3.9K]3 years ago
3 0
The elements present in Ammonium Nitrate are Hydrogen, Nitrogen, and Oxygen at a ratio of 4:2:3, respectively. Hydrogen weighs in at 1.008 amu, Nitrogen at 14.007, and Oxygen at 15.999. This means that the molar mass would be:

Hydrogen
4 x 1.008 = 4.032 amu

Nitrogen
2 x 14.007 = 28.014 amu

Oxygen
3 x 15.999 = 47.997 amu

Total
4.032 + 28.014 + 47.997 = 80.043 amu

The molar mass of Ammonium Nitrate is 80.043 grams per mole.
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In increasing order:

Sr, Ca, Mg, Br

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Why krypton and noble gas are not used in airships using the periodic table?
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Since the noble gases are unreactive or inert, they are safe to use. Helium is used to fill balloons and airships, because it is much lighter that air and it will not catch fire. Neon is used in advertising signs. It will give red glow, but the color can be changes by mixing it with other gases.

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Brainliest for correct answer please show all work
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Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

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To balance an equation, ensure that the number of atoms of each element is equal on both sides.

Reactants would be those on the left of the arrow while products are on the right of the arrow.

Balance O and H atoms last.

<u>Question 1:</u>

__Na₃PO₄ + __KOH ➙ __NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 1Na, 1P, 3K, 5O, 1H

<em>Balance the number of Na:</em>

__Na₃PO₄ + __KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 1K, 5O, 1H

Products: 3Na, 1P, 3K, 7O, 3H

<em>Balance the number of K:</em>

__Na₃PO₄ + 3 KOH ➙ 3 NaOH + __K₃PO₄

Reactants: 3Na, 1P, 3K, 7O, 3H

Products: 3Na, 1P, 3K, 7O, 3H

<em>The equation is now balanced.</em>

<u>Question 2:</u>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + __LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 1F, 1Li, 1C, 3O

<em>Balance</em><em> </em><em>n</em><em>u</em><em>m</em><em>b</em><em>e</em><em>r</em><em> </em><em>of</em><em> </em><em>L</em><em>i</em><em> </em><em>and</em><em> </em><em>F</em><em> </em><em>atoms</em><em>:</em>

__MgF₂ + __Li₂CO₃➙ __MgCO₃ + 2 LiF

Reactants: 1Mg, 2F, 2Li, 1C, 3O

Products: 1Mg, 2F, 2Li, 1C, 3O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

<u>Question 3:</u>

__P₄ + __O₂➙ __P₂O₃

Reactants: 4P, 2O

Products: 2P, 3O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>P</em><em> </em><em>atoms</em><em>:</em>

__P₄ + __O₂➙ 2 P₂O₃

Reactants: 4P, 2O

Products: 4P, 6O

<em>Balance</em><em> </em><em>the</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>O</em><em> </em><em>atoms</em><em>:</em>

__P₄ + 3 O₂➙ 2 P₂O₃

Reactants: 4P, 6O

Products: 4P, 6O

<em>The</em><em> </em><em>equation</em><em> </em><em>is</em><em> </em><em>now</em><em> </em><em>balanced</em><em>.</em>

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