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I am Lyosha [343]
3 years ago
14

A sample of the compound magnesium oxide is synthesized as follows. 60. g of magnesium is burned and produces 100. g of magnesiu

m oxide, indicating that the magnesium combined with 40. g of oxygen in the air. If 30. g of magnesium had been used, what mass of oxygen would have combined with it? What law of chemistry is used in solving this problem?
Chemistry
1 answer:
Inga [223]3 years ago
5 0

Answer:

30 g of magnesium would be combined with 20 g of oxygen. The law used solving this problem is the Lavoisier Law of conservation of mass.

Explanation:

If 60 g of magnesium combines with 40 g of oxygen to make 100 g of magnesium oxide, then 30 g of magnesium will combine with 20 g of oxygen to make 50 g of magnesium oxide.

This happens because in a chemical reaction there is no creation or descruction of atoms, only a rearrangement. Therefore, the mass of reactants should be equal to the mass of products.

The following equation represents the proportions of the substances:

Mg + 1/2O₂ → MgO

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The gram formula mass of a compound is 48 grams. The mass of 1.0 moles of this compound is
ehidna [41]
The gram formula mass is Molar mass. The mass of 1.0 moles is : 
3) 48.0 g
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3 years ago
How many moles of KF would need to be added to 2500 ml of water to make 1.2 M solution?
Oduvanchick [21]

The number of moles of KF needed to prepare the solution is 3 moles

<h3>What is molarity?</h3>

Molarity is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

<h3>How to determine the mole of KF </h3>
  • Volume = 2500 mL = 2500 / 1000 = 2.5 L
  • Molarity = 1.2 M
  • Mole of KF =?

Molarity = mole / Volume

1.2 = mole of KF / 2.5

Cross multiply

Mole of KF = 1.2 × 2.5

Mole of KF = 3 moles

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4 0
2 years ago
Given the following reaction: 3CuCl2(aq) 2Na3PO4(aq) --&gt; Cu3(PO4)2(s) 6NaCl(aq) MM (g/mol) 134.45 163.94 380.58 58.44 If 285
nata0808 [166]

Answer:

227.78g of the precipitate are produced

Explanation:

Based on the reaction, 3 moles of CuCl2 produce 1 mole of Cu3(PO4)2 (The precipitate).

To solve this question we need to find the moles of CuCl2 added. With these moles and the reactio we can find the moles of Cu3(PO4)2 and its mass as follows:

<em>Moles CuCl2:</em>

285mL = 0.285L * (6.3mol / L) = 1.7955 moles CuCl2

<em>Moles Cu3(PO4)2:</em>

1.7955 moles CuCl2 * (1mol Cu3(PO4)2 / 3mol CuCl2) = 0.5985 moles Cu3(PO4)2

<em>Mass Cu3(PO4)2 -380.58g/mol-</em>

0.5985 moles Cu3(PO4)2 * (380.58g/mol) =

227.78g of the precipitate are produced

7 0
3 years ago
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