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MatroZZZ [7]
3 years ago
15

2. 10.00 grams of a sample of hydrated PtCl4 are heated and lose 3.00 g of water. How many moles of water are combined with each

mole of PtCl4?
Chemistry
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

8 mol

Explanation:

Step 1: Calculate the mass of PtCl₄ in the sample

10.00 grams of a sample of hydrated PtCl₄ are heated and lose 3.00 g of water. The mass of PtCl₄ is:

mPtCl₄ = 10.00 g - 3.00 g = 7.00 g

Step 2: Calculate the moles corresponding to 7.00 g of PtCl₄ and 3.00 g of H₂O

The molar mass of PtCl₄ is 336.9 g/mol.

7.00 g × 1 mol/336.9 g = 0.0208 mol

The molar mass of H₂O is 18.02 g/mol.

3.00 g × 1 mol/18.02 g = 0.166 mol

The molar ratio of H₂O to PtCl₄ is:

0.166 mol H₂O/0.0208 mol PtCl₄ ≈ 8 mol H₂O/ 1 mol PtCl₄

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Explanation:

Hardness of rocks makes weathering difficult and not easy to come by.

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The agents of weathering are primarily the agents of denudation which are wind, rainfall, glacier, gravity and temperature.

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6 0
4 years ago
What’s is the percent composition by mass of oxygen in magnesium oxide , Mg0
Advocard [28]

Answer:

40%

Explanation:

We'll begin by obtaining the molar mass of MgO. This is illustrated below:

Molar Mass of MgO = 24 + 16 = 40g/mol

Observing the formula MgO, we have 1 atom of O in it.

The percentage composition by mass of oxygen in MgO is given by:

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6 0
3 years ago
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Writ
Nesterboy [21]

Answer:

2ErF3 +3Mg = 3MgF2 + 2Er

Explanation:

This is a single replacement equation where there are 2 metals. The bonds are broken and new bonds are formed again by Mg and F.

Er has a +3 charge and F has a -1 charge. You switch it around and you get ErF3. Then you add the second reactant, Mg. The product is MgF as stated, and Mg has a charge of +2 and F has -1. You switch it again and you get MgF2. Then the second product Er is there.

Now we have

ErF3+Mg=MgF2+Er

So we balance the equation because of the law of conservation of mass.

Make F equal, so we add the coefficents 2 and 3

2ErF3+Mg=3MgF2+Er

And now Mg and Er need balancing so

2ErF3+3Mg=3MgF2+2Er

Hope this helped

8 0
3 years ago
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The correct answer is E.
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