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Finger [1]
3 years ago
9

Which of the following compounds contains both ionic and covalent bonds? KOH N2O5 CH3OH Na2O

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

Answer: a.KOH

Potassium hydroxide is an ionic compound where the K+ is the cation and OH−is the anion. At the same time, the compound also contains a covalent bond since the anion, OH−is formed from electron sharing between the O and H atoms.

Hope this helps........ Stay safe and have a Merry Christmas! :D

Explanation:

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5000kg of ammonium nitrate per square kilometer of cornfield per year. how much nitric acid would be needed to make the fertiliz
Serggg [28]

NH_3+HNO_3-> NH_4NO_3

1 mole of nitric acid produce 1 mole of ammonium nitrate.

moles in 5000 kg of ammonium nitrate :

n=\dfrac{5000\times 1000}{80}\\\\n=62500\ moles ( molecular mass of ammonium nitrate is 80 gm/mol )

So, number of moles of nitric acid required are also 62500 moles.

Mass of 62500 moles of nitric acid :

mass = 62500\times 63 \\\\mass = 3937500\ gram\\\\mass = 393.75\ kg

Hence, this is the required solution.

5 0
2 years ago
How many grams of fluorine are contained in 8 molecules of boron trifluoride?
Lelu [443]
<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

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B: quantity of solute with a specific volome of solvent.

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