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nignag [31]
3 years ago
14

How many valence electrons does the Lewis structure for a chlorine atom show

Chemistry
2 answers:
Marianna [84]3 years ago
8 0

7 valance electrons the Lewis structure for a chlorine atom show.

<u>Explanation:</u>

Lewis structure is used to keep track of the valance electrons of each atom and share them in bonding. Every valance electron is represented as a dot around the element to keep their count.  

Paired VE’s are shown as ‘:’ and unpaired as ‘.’ in the representation. A chlorine Cl atom has 17 electrons. & of them are valance electrons. Out of the seven six are paired electrons and one is unpaired.  

ohaa [14]3 years ago
6 0

Answer: D

Explanation:

Chlorine is in group 7 or (VII) in Roman numerals, which means it has 7 balance electrons. It only needs one electron to become stable, hence it is next to the noble gases

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What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not
Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 =  [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

7 0
3 years ago
1‑propanol ( n ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure a
Alex777 [14]

Answer:

y_{prop} = 0.134; y_{iso} = 0.866

The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr

Explanation:

For each of the solutions:

mole fraction of isopropanol  (x_{iso}) = 1 - mole fraction of propanol (x_{prop}).

Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.

Furthermole, the partial pressure of isopropanol = x_{iso}*vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr

The partial pressure of propanol = x_{prop}*vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr

Similarly,

In the vapor phase,

The mole fraction of propanol (y_{prop}) = \frac{P_{prop} }{P_{prop}+P_{iso}}

Where, P_{prop} is the partial pressure of propanol and P_{iso} is the partial pressure of isopropanol.

Therefore,

y_{prop} = 5.26/(34.04+5.16) = 0.134

y_{iso} = 1 - 0.134 = 0.866

5 0
3 years ago
How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?
mars1129 [50]

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

3 0
2 years ago
What is the molar mass of just the O in CO2?
Vlad1618 [11]

Answer:

16 g/mol

Explanation:

In CO2, it means we have 1 mole of carbon and 2 moles of oxygen.

However, we want to find the molar mass of just a single mole of oxygen.

Now, from tables of values of elements in electronic configuration, the molar mass of oxygen is usually approximately 16 g/mol.

In essence the molar mass is simply the atomic mass in g/mol

8 0
3 years ago
A drinking water plant adds 500 grams of fluoride to a water tank containing 500,000 liters of drinking water. what is the conce
nlexa [21]
Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.
5 0
2 years ago
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