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2 years ago

How many electron energy shells are occupied in an unstable atom of silicon

Chemistry

1 answer:

drek231 [11]2 years ago

8 0

Answer:

Explanation:

it will share its valence electrons with other elements to acquire noble gas confriguration of the nearest inert element.

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How many moles of water in 100 grams

Bond [772]

Answer:

There are approximately 5.55 moles

Explanation:

6 0

2 years ago

A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask

mr_godi [17]

Answer:

The concentration in mol/L = 4.342 mol/L

Explanation:

Given that :

mass of sodium chloride = 25.4 grams

Volume of the volumetric flask = 100 mL

We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol

and number of moles = mass/molar mass

The number of moles of sodium chloride = 25.4 g/58.5 g/mol

The number of moles of sodium chloride = 0.434188 mol

The concentration in mol/L = number of mol/ volume of the solution

The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L

The concentration in mol/L = 4.34188 mol/L

The concentration in mol/L = 4.342 mol/L

5 0

3 years ago

How is a water wave similar to a pendulum?

Flura [38]

Answer:

side to side?

Explanation:

3 0

2 years ago

How many significant figures are in the following number? 5.67 x 106<br> 3<br> 4<br> 5<br> 6

mixas84 [53]

3 these are 5,6&7 are the significant

6 0

3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

$MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}$

Thus, we can conclude that half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

3 0

3 years ago