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vazorg [7]
2 years ago
5

How many protons and electrons does oxygen have? What about hydrogen? Draw a model of each atom if that helps you think about th

e differences between them ?
Chemistry
1 answer:
Tems11 [23]2 years ago
5 0

Answer:

Two electrons fit in the first shell out from the nucleus and eight fit in the second. Every element with more protons than the two of Helium needs to work on shells outside the first one. one you get to ten, you have filled the first two shells.

In a water molecule, oxygen forms one covalent bond with EACH of TWO hydrogen atoms. As a result, the oxygen atom has a stable arrangement of 8 valence electrons. Each hydrogen atom forms only one bond because it needs only two electrons to be stable.

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Is anyone good at chemistry if so can someone help me please ?<br><br> (NO LINKS)
Mkey [24]

This requires familiarity with the different theories (or concepts) of acids and bases.

On the Arrhenius concept, an acid is a substance that produces an H⁺ ion in water such that the H⁺ concentration increases, and a base is a substance that produces an OH⁻ ion in water such that the OH⁻ concentration increases.

On the Brønsted–Lowry concept, an acid is a substance that donates a proton (which is basically an H⁺ ion) in a solvent, and a base is a substance that accepts a proton in a solvent.

On the Lewis concept, an acid is a substance that accepts an electron pair in a solvent, and a base is a substance that donates an electron pair in a solvent.

The concepts become progressively broader, i.e., the Arrhenius concept is the most restrictive and the Lewis concept is the least restrictive. As a corollary, an Arrhenius acid or base is also both a Brønsted–Lowry acid or base and a Lewis acid or base, respectively; a Brønsted–Lowry acid or base is not necessarily an Arrhenius acid or base, but an Arrhenius acid or base is also a Lewis acid or base, respectively. And finally, a Lewis acid or base may not necessarily be either an Arrhenius or a Brønsted–Lowry acid or base.

So, with the above concepts in mind, we can match the statements in column A with the type of acid or base in column B:

\begin{center}\begin{tabular}{ c c } 1 & Bronsted Lowry acid \\  2 & Bronsted Lowry base \\   3 & Arrhenius acid \\ 4 & Arrhenius base \\ 5 & Lewis base \\ 6 & Lewis acid\end{tabular}\end{center}

6 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
Most chloride salts are soluble. Which of the following is an exception to this generalization?
siniylev [52]

The exception to the rule concerning the solubility of chlorides in water is PbCl2.

The solubility rules give us an idea of which substances are soluble in water and what substances are not soluble in water. According to the solubility rules, chlorides are soluble in water.

However, chlorides of lead are not soluble in water hence, the exception to the rule is PbCl2.

Learn more: brainly.com/question/6505878

7 0
3 years ago
A lactic acid/lactate ion buffer solution contains 0.14 M HC3H5O3 and 0.83 M C3H5O3−, respectively. The Ka value of lactic acid
yKpoI14uk [10]
Use Henderson Hasselbache
PH=pka+log([salt]/[acid])
Pka=-log(ka)
-log(1.4x10^-4)= 3.85
log(.83/.14)= .77
3.85+.77= 4.62
PH=4.62
6 0
3 years ago
Read 2 more answers
What is the volume of 0.80 grams of O2 gas at STP?
frez [133]

Answer:

The volume of 0.80 grams of O2 is 0.56 L

Explanation:

Step 1 : Data given

Mass of O2 = 0.80 grams

Molar mass of O2 = 32 g/mol

STP = 1 mol, 1atm, 22.4 L

Step 2: Calculate moles of oxygen

Moles of O2 = Mass of O2 / molar mass of O2

Moles O2 = 0.80 grams / 32 g/mol

Moles O2 = 0.025 moles

Step 3: Calculate volume of O2

1 mol = 22.4 L of gas

0.025 moles = 0.025*22.4L = 0.56 L

The volume of 0.80 grams of O2 is 0.56 L

5 0
4 years ago
Read 2 more answers
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