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nekit [7.7K]
3 years ago
13

Why is excess magnesium oxide used in the formation of magnesium sulfate crystals?​

Chemistry
2 answers:
Tom [10]3 years ago
3 0

I really don't know what the answer is

Anestetic [448]3 years ago
3 0

Answer:

When you want magnesium sulfate salts in lab use magnesium oxide and magnesium carbonate to prepare magnesium sulfate,because magnesume react with sulfuric acids to produce hydrogen gas and that is very dangerous because it explode easily.

Explanation:

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Leya [2.2K]

Answer:

It’s B just did it

Explanation:

6 0
3 years ago
What are the blanks the missing blanks?
mixas84 [53]

Answer:

Cr: Mass- 51.9961 u

Fe: Moles- 55.845

Ti: Mass- 47.867 u

Hg: Moles- 200.59.

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6 0
3 years ago
Which of the following statements about elements is not true?
guapka [62]

Answer:

Option A is not true

Explanation:

Could you please follow me and mark me as the brainliest answer

6 0
3 years ago
Brainliest for correct answer please show all work
Korvikt [17]

Answer:

1) Na₃PO₄ + 3 KOH ➙ 3 NaOH + K₃PO₄

2) MgF₂ + Li₂CO₃➙ MgCO₃ + 2 LiF

3) P₄ + 3 O₂➙ 2 P₂O₃

Explanation:

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>

4 0
3 years ago
What is the chemical equation for the alpha decay of erbium-144
iren [92.7K]

Answer:

Er-144 -------> Dy-140 + He-4

Explanation:

Alpha decay is the release of a hydrogen nucleus.  So the original atom will decrease the mass by 4 and the atomic number by 2.

8 0
3 years ago
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