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Alona [7]
3 years ago
9

Identify the salt that is produced from the acid-base neutralization reaction between potassium hydroxide and acetic acid.

Chemistry
1 answer:
schepotkina [342]3 years ago
4 0
Potassium Acetate .
CH3COOK
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Which fact below is describing a base?
QveST [7]
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6 0
3 years ago
For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.
Diano4ka-milaya [45]
The balanced chemical reaction is given as follows:

<span>2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The starting amount of the reactant are given above. These values would be used for the calculations. We do as follows:

</span>2.72 g KClO3 (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 1.06 g O2
<span>
0.361 g KClO3 </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.14 g O2
<span>
83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
<span>
22.5 mg KClO3</span> (1 g / 1000 mg) (1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 0.009 g O2
4 0
2 years ago
Read 2 more answers
What is the unabbreviated electron configuration for Neptunium?
slava [35]

Answer:

Rn 5f4 6d1 7s2

Explanation:

6 0
2 years ago
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HELP PLEASE!! ASAP What would the mass be in kg?
fgiga [73]

Answer:

i guess answer is 0.0600

Explanation:

Here,

Density=0.791g/cm^3

Volume=75.89ml

So,

Mass=Density*Volume

=0.791g/cm^3×75.89ml

=60.02g/cm^3.ml

Expressing them in kg

60.02/1000

=0.0600 kg

4 0
2 years ago
When solutions of silver nitrate and sodium chloride are mixed, a precipitation reaction occurs. What mass of precipitate can be
Bas_tet [7]

Explanation:

The given precipitation reaction will be as follows.

        AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)

Here, AgCl is the precipitate which is formed.

It is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity = \frac{\text{no. of moles}}{\text{volume in liter}}

It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.

                    Molarity = \frac{\text{no. of moles}}{\text{volume in liter}}

                     0.269 M = \frac{\text{no. of moles}}{1.14 L}

                    no. of moles = 0.306 mol

As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.

               No. of moles = \frac{mass}{\text{molar mass}}

                   0.307 mol = \frac{mass}{143.32 g/mol}

                           mass = 43.99 g

Thus, we can conclude that mass of precipitate produced is 43.99 g.

                 

8 0
2 years ago
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