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bogdanovich [222]
3 years ago
11

Elements bond when they share or transfer their outer shell electrons which are called valence electrons

Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

True

Explanation:

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Plant cells and animal cells were observed under a microscope. The characteristics of two cells are listed below.
IRINA_888 [86]

Answer:

The answer is B

Explanation:

ITs B

4 0
3 years ago
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What are the relevant equations for oxidation and reduction<br><br>​
kakasveta [241]

Answer:

There are many equations for oxidation and reduction, but the most relevant ones are the equations for the half-reactions. In general, the equation for oxidation is:

Oxidation: Reducing Agent + O2 → Oxidizing Agent

And the equation for reduction is:

Reduction: Oxidizing Agent + e- → Reducing Agent

These equations show the transfer of electrons that occurs during oxidation and reduction.

Explanation:

7 0
2 years ago
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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.

        <u>      Number of atoms             Electrons to achieve Octet       Needed</u>

Si                      1                      x                         8                           =       8

Cl                     4                      x                         8                           = <u>     32</u>

                                                                     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.

        <u>      Number of atoms             Valence electrons                 Available</u>

Si                      1                      x                         4                           =      4

Cl                     4                      x                         7                          = <u>     28</u>

                                                                     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 mass of NaC6H5COO should be added to 1.5 L of 0.40 M C6H5COOH solution at 25 °C to produce a solution with a pH of 3.87 giv
statuscvo [17]

Answer:

41 g

Explanation:

We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.

pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]

pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]

log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]

log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40

[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M

We can find the mass of NaC₆H₅COO using the following expression.

M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution

mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L

mass NaC₆H₅COO = 41 g

7 0
3 years ago
Which factor determine the reactivity of alkyl halide
Pavel [41]

Answer:

This reactivity order reflects both the strength of the C–X bond, and the stability of X(–) as a leaving group, and leads to the general conclusion that alkyl iodides are the most reactive members of this functional class.

6 0
3 years ago
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