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3 years ago

Fluorine has the electron configuration 1s22s22p5. With which of the following elements would fluorine form an ionic bond?

Chemistry

1 answer:

Rashid [163]3 years ago

5 0

Fluorine since it is a halogen (non-metal), then it would need to bond with a metal in order to form an ionic bond.

Fluorine would therefore need an element from any of the metal groups to bond with in order to form an ionic compound e.g. Group One (Sodium or Lithium) or Group Two (Magnesium or Calcium) or Group Three (Aluminum) or a Transitional metal (Copper or Vanadium).

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3 years ago

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If the pH is 4.5, what is the pOH ?

Ahat [919]

PH+pOH=14
4.5+pOH=14
pOH=14-4.5
pOH = 9.5

We know that
[H⁺][OH⁻]=10⁻¹⁴,so

-log([H⁺][OH⁻])=-log10⁻¹⁴
-log [H⁺]+ (- log[OH⁻])= - log10⁻¹⁴,
- log[H⁺] = pH
- log[OH⁻] = pOH
- log10⁻¹⁴=-(-14)log10=14*1=14
pH + pOH =14

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3 years ago

Which of the following is a structure of a cow?

AfilCa [17]

Answer:

None of the above.

Explanation:

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3 years ago

When a student adds benedict’s solution to a clear solution, the solution turns orange. the student can conclude that the soluti

4vir4ik [10]

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3 years ago

A 0.9440 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

Gnom [1K]

Answer : The percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.9440 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.9440-x)g}{74.5g/mole}=\frac{(0.9440-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.9440-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{1.903g}{143.32g/mole}=0.0133moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

0.0133 mole = $\frac{x}{58.5}moles+\frac{(0.9440-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.171g$

The mass of NaCl = x = 0.171 g

The mass of KCl = (0.9440 - x) = 0.9440 - 0.171 = 0.773 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.171g}{0.9440g}\times 100=18.11\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.773g}{0.9440g}\times 100=81.88\%$

Therefore, the percent by mass of NaCl and KCl are, 18.11 % and 81.88 % respectively.

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4 years ago