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lord [1]
3 years ago
9

28. How many formula units are found in 0.250 moles of potassium nitrate?

Chemistry
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

\boxed {\boxed {\sf B. \ 1.5 *10^{23} \ formula \ units}}

Explanation:

1 mole of any substance contains the same number of particles. The particles can vary (atoms, molecules, formula units), but there are always 6.022*10²³ particles. In this case, the particles are formula units of potassium nitrate or KNO₃.

Let's create a ratio.

\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 \ mol \ KNO_3}

Since we are trying to find the formula units in 0.250 moles, we multiply by that number.

0.250 \ mol \ KNO_3 *\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 \ mol \ KNO_3}

The units of moles of potassium nitrate cancel.

0.250  *\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 }

The denominator of 1 can be ignored, so we can make a simple multiplication problem.

0.250  *{6.022*10^{23} \ formula \ units  \ KNO_3}

1.5055 * 10^{23} \ formula \ units \ KNO_3

If we round to the nearest tenth, the 0 in the hundredth place tells us to leave the 5 in the tenth place.

1.5 *10^{23} \ formula \ units \ KNO_3

0.250 moles of potassium nitrate is approximately equal to 1.5*10²³ formula units of potassium nitrate and choice B is correct.

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4 0
3 years ago
Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction
11Alexandr11 [23.1K]

Answer:

<u>Balanced equation:</u>

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

<u>Ionic equation:</u>

Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)

4 0
3 years ago
Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

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labwork [276]

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I hope this is helpful

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Hello.

The answer is: D it produces hyrogen ions in a solution.

This is correct because when Arrhenius acid it turns into hydrogen ions.substance as an acid if it produces hydrogen ions H(+) or hydronium ions in water. A substance is classified as a base if it produces hydroxide ions OH(-) in water.

have a nice day
3 0
3 years ago
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