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3 years ago

What is the formula for Na+1 and SO3^-2

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer:

Na₂SO₃

Explanation:

We can use the criss-cross method to work out the formula.

1. Write the symbols of the anion and cation.

Na¹⁺ SO₃²⁻

2. Criss-cross the numbers of the charges to become the subscripts of the other ion (see image).

3. Write the formula with the new subscripts.

Na₂(SO₃)₁

4. Omit all subscripts that are 1.

The formula becomes Na₂SO₃.

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5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(

professor190 [17]

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

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4 years ago

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What is the solubility of Na2HAsO4 at 60 degrees C

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C. 65g / 100ml of water

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3 years ago

Changing a solid to a liquid to a gas is changing the matter. * Your answer

sukhopar [10]

Answer:

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Explanation:

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3 years ago

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How does matter change from one state to another?

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3 years ago

How many moles of oxygen are needed for the complete combustion of 29.2 grams of acetylene?

podryga [215]

Moles of Oxygen= 2.8075 moles

<h3>Further explanation</h3>

Given

29.2 grams of acetylene

Required

moles of Oxygen

Solution

Reaction(Combustion of Acetylene) :

2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)

Mol of Acetylene :

= mass : MW Acetylene

= 29.2 g : 26 g/mol

= 1.123

From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :

= 5/2 x mol C₂H₂

= 5/2 x 1.123

= 2.8075 moles

7 0

3 years ago