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taurus [48]
3 years ago
7

Draw the atomic structure with electronic configuration of Potassium​

Chemistry
2 answers:
DedPeter [7]3 years ago
4 0

Answer:

Explanation:

i try to help you

bezimeni [28]3 years ago
3 0

✧Kindly see the attached picture! ☄

♪ Hope I helped! ♡

\star Best regards! ツ

☞ \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}} ✎

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Calculate the volume of a 2.0 molar aqueous solution made from 14 miles of K2S
Vilka [71]

Answer:

volume is 7.0 liters

Explanation:

We are given;

  • Molarity of the aqueous solution as 2.0 M
  • Moles of the solute, K₂S as 14 moles

We are required to determine the volume of the solution;

We need to know that;

Molarity = Moles ÷ volume

Therefore;

Volume = Moles ÷ Molarity

Thus;

Volume of the solution = 14 moles ÷ 2.0 M

                                       = 7.0 L

Hence, the volume of the molar solution is 7.0 L

5 0
3 years ago
In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il
Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.        

7 0
3 years ago
Read 2 more answers
Will give brainliest! if 32.0 g of hcl is to be diluted to make a 4.80 m solution, how much water should be added?
nadezda [96]

The required volume of water is 0.18 liters.

<h3>What is molarity?</h3>

Molarity of any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V

Moles of solute will be calculated as:

n = W/M, where

W = given mass of HCl = 32g

M = molar mass of HCl = 36.4g/mol

n = 32 / 36.4 = 0.88 mole

Given molarity of solution = 4.80M

On putting all values in the above equation, we get

V = (0.88) / (36.4) = 0.18 L

Hence required volume of water is 0.18L.

To know more about volume & concentration, visit the below link:

brainly.com/question/26762947

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6 0
2 years ago
When a candle burns what form of energy does the chemical energy in the candle change
soldier1979 [14.2K]
The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs, and produces heat and light.
8 0
3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
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