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valina [46]
4 years ago
15

-- A scientist combines 14 grams of potassium nitrate with an unknown amount of

Chemistry
1 answer:
igor_vitrenko [27]4 years ago
5 0

Answer:

The unknown amount of potassium chloride is 13.6 grams.            

 

Explanation:

The reaction of 14 grams of KNO₃ with KCl produces a total mass of 27.6 grams of the products.

The law of conservation of mass tells us that the total mass of the reactants must be the same that the total mass of the products. So, we can find the mass of KCl as follows:

m_{r} = m_{p}

Where <em>r</em> is for reactants and <em>p </em>is for products

m_{KNO_{3}} + m_{KCl} = m_{p}

14 g + m_{KCl} = 27.6 g

m_{KCl} = 27.6 g - 14 g = 13.6 g

Therefore, the unknown amount of potassium chloride is 13.6 grams.           

I hope it helps you!

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Answer:

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Explanation:

A <em>positive entropy change</em> means that the entropy of the products is greater than the entropy of the reactants.

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Since 5 molecules of a gas (high disorder) combines with 2 molecules of liquid to produce 4 units of aqueous HNO₃ you may expect that the product is more ordered than the reactants, which means that the change in entropy is negative (the entropy decreases).

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<u>B) Fe₂O₃ (s) + 3C(s) → 2Fe(s) + 3CO₂(g)</u>

The left side (reactants) show only solid substances which is a highly ordered arrangement while the right side (products) show the formation a solid (ordered arrangement) and a gas (highly disoredered arrangement), so you can predict the increase of the system entropy, i.e. a positive entropy change.

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The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = \frac{m (mass)}{ v (volume)}. Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:

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Answer:

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