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Zigmanuir [339]
2 years ago
8

The atomic number of fluorine is 9. How many electrons does an ion of fluorine have if it is represented by the symbol shown bel

ow?
19F1-

A. 8 electrons
B. 9 electrons
C. 10 electrons
D. 20 electrons
Chemistry
2 answers:
Orlov [11]2 years ago
8 0

Answer:

b. 8 electrons

Explanation:

hope it's hepl

Irina-Kira [14]2 years ago
5 0
  • See charge on ion is -1 .

Hence it has taken 1 electron

Now first look at EC of Fluorine(F)

\\ \bull\sf\dashrightarrow 1s^22s^22p^5

  • Now one electron added .hence no of electrons is 10now

Look at the EC

\\ \bull\sf\dashrightarrow 1s^22s^22p^6

Or

\\ \bull\sf\dashrightarrow [He]

Option C is correct.

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What is the pH of a 0.083 solution of HCI
quester [9]

Answer:

The pH is 1.08,  being an acid pH

Explanation:

pH is a parameter used to measure the degree of acidity or alkalinity of a substance. The pH scale ranges from 0 to 14. Values ​​less than 7 indicate the acidity range and those greater than 7 indicate alkalinity or basicity. Value 7 is considered neutral. Mathematically, pH is the negative logarithm of the molar concentration of hydrogen or proton ions (H⁺) or hydronium ions (H₃O⁺).

pH= - log [H⁺]= - log [H₃O⁺]

A strong acid is an acid that completely dissociates into hydrogen ions and anions in solution. This implies that the initial concentration of acid is equal to the final concentration of H₃O⁺. This occurs with acid HCl.

So,  pH is calculated as:

pH= - log (0.083)

pH= 1.08

<u><em>The pH is 1.08,  being an acid pH</em></u>

8 0
3 years ago
A sample of nitrogen gas has a volume of 578 mL and a pressure of 124.1 kPa. What volume would the gas occupy at 80.2 kPa if the
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Answer:

V2 = 894.4mL

Explanation:

P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?

Applying Boyle's law

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124.1*578=80.2*V2

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4 0
3 years ago
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
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The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

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Answer:

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