Can violance be justified when words fail to bring social justice

I don’t understand how to find these please help

ddd [48]

Answer:

8 shared electrons

Explanation:

When you are looking for the number of shared electrons via the equation:

S = N-A

Where:

S = means the shared electrons

N = Needed electrons

A = available electrons

"Needed electrons" means how many electrons does it need to have a noble gas configuration, in this case, to complete the octet rule.

"Available electrons" means how many valence electrons is actually available considering the compound or the elements involved in the compound.

To get the needed electrons, treat the elements involved separately. We have a silicon (Si) atom and 4 chlorine (chlorine) atoms in this compound. Let's list it down first:

Number of atoms

Si 1

Cl 4

Next step is to determine how many electrons should it have in its outer shell to achieve the octet rule. Both of them in this case would be 8. Multiply that by the number of atoms and add up the needed electrons to determine how many you will need for this particular compound.

Number of atoms Electrons to achieve Octet Needed

Si 1 x 8 = 8

Cl 4 x 8 = 32

TOTAL: 40

This is now our N. N = 40 electrons

Next step is to determine how many we actually have. Your clue in determining how many valence electrons the atom has is the group. Silicon is in Group 4A, this means it has 4 valence electrons. Chlorine is in Group7A, so ths means it has 7 valence electrons.

So first we write the number of atoms again, then in the next column, you write down the actual number of valence electrons and multiply them. Sum it up to see how many electrons available in this particular compound.

Number of atoms Valence electrons Available

Si 1 x 4 = 4

Cl 4 x 7 = 28

TOTAL: 32

This is now our A. A = 32 electrons

Now we apply this:

S = N - A

N = 40 electrons

A = 32 electrons

S = 40 - 32 = 8

Number of shared electrons is 8

5 0

3 years ago

What are the functional groups in oxybenzone?

slava [35]

Answer:

Oxybenzone contains ether, phenol and ketone (-CO) functional group alonq with two aromatic rings.

Explanation:

5 0

4 years ago

Read 2 more answers

How many valence electrons are in an atom of magnesium?

dimulka [17.4K]

There are two valence electrons in a single atom of magnesium.

8 0

3 years ago

Read 2 more answers

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

What phenomenon does the Coriolis effect arise from?

Andre45 [30]

Answer:

because our planet is spinning

which means the objects near the equator are moving at much faster velocities than objects at higher latitudes

I hope this helps✌

7 0

3 years ago