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3 years ago

Which element will only gain one electron during a chemical reaction?

Chemistry

1 answer:

Gelneren [198K]3 years ago

7 0

Halogens

Explanation:

Halogens are a group of non-metals located in the seventh group on the periodic table. The will only gain one electron during a chemical reaction.

Halogens have a seven electrons in their outermost shell.
To complete the number of electrons in this shell, they need to gain an additional electron.
One more electron makes the halogen similar to the corresponding noble gas which is very stable.
Halogens are very reactive groups of elements and are highly electronegative.
They have a high affinity for electrons.
These elements are fluorine, chlorine, bromine, iodine and Astatine.

learn more:

Halogens brainly.com/question/6324347

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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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