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asambeis [7]
3 years ago
8

A compound composed of nickel and fluorine contains 9.11 g ni and 5.89 g f. what is the empirical formula of this compound

Chemistry
1 answer:
aalyn [17]3 years ago
6 0
Empirical formula is the simplest formula showing the simplest ratio of atoms in a compound. Calculated as shown;
 we start by calculating the number of moles of each atom;
moles of nickel = 9.11 g ÷ 58.7 g = 0.155 moles 
moles of fluorine = 5.89 g ÷ 19 g = 0.31 moles
Then we get the ratio of the moles of nickel to that of flourine
 That is 0.155 : 0.31  (dividing by the smallest)
            0.155/0.155 : 0.31/0.155
we get 1:2  ( the simplest ratio)
Therefore the empirical formula is nif2 
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Examples of physical and chemical weathering
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3 years ago
What is the mass of 3.6 x 1024 atoms of Zinc (Zn)?
AlladinOne [14]
Answer is: mass od zinc is 392,28 g.
N(Zn) = 3,6·10²⁴.
n(Zn) = N(Zn) ÷ Na.
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7 0
3 years ago
A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
Igoryamba

Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

  • Unsaturated: the concentration is below the maximum concentration permited at the given temperature.

  • Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.

  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

<em />

3 0
3 years ago
If 20.0 g LIOH react with excess KCl, giving a 17.0
adell [148]

The yield of lithium chloride is 1.92 grams.

Option D.

<h3><u>Explanation:</u></h3>

In this reaction, we can see that 1 mole of lithium hydroxide reacts with 1 mole of potassium chloride to produce 1 mole of lithium chloride and 1 mole of potassium hydroxide.

Molecular weight of lithium hydroxide is 24.

Molecular weight of lithium chloride is 42.5.

So 24 grams of lithium hydroxide produces 42.5 grams of lithium chloride.

So, 20 grams of lithium hydroxide produces 20 * 24 / 42.5  grams =11. 29 grams of lithium chloride.

But this is when the yield is 100%.

But yield is 17%.

So the yield is 1.92 grams of lithium chloride.

8 0
3 years ago
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