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mezya [45]
3 years ago
10

When at atom is charged, it is called an ion. how many electrons are in o2?

Chemistry
2 answers:
fenix001 [56]3 years ago
5 0
32

Electrons are the same number as the protons unless it gives you a charge of +1

So O has 16 then multiply it by 2 and that gives you an answer of 32.

Hope this helps
Nina [5.8K]3 years ago
4 0
10 electrons are in O2. 
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What occurs to the atoms of reactants in a chemical reaction?

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The law of definite proportions states that______.
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Answer:

Law of definite proportions, statement that every chemical compound contains fixed and constant proportions (by mass) of its constituent elements.

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3 years ago
What is the volume of a balloon of gas at 842 mm Hg and -23° C, if its volume is 915 mL at a pressure of 1,170 mm Hg and a tempe
garik1379 [7]
Answer:
             V₂  =  1070 mL or 1.07 L

Solution:

Data Given;
                  P₁  =  1170 mmHg

                  V₁  =  915 mL

                  T₁  =  24 °C  +  273 K  =  297 K

                  P₂  =  842 mmHg

                  V₂  =  ?

                  T₂  =  - 23 °C  +  273 K  =  250 K

According to Ideal gas equation,

                       P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                       V₂  =  P₁ V₁ T₂ / P₂ T₁

Putting Values,

                       V₂  = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)

                       V₂  =  1070 mL or 1.07 L
5 0
4 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.
Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = pK_{a} + log \frac{Base}{Acid}  

          4.245 = 3.75 + log \frac{Base}{Acid}

      log \frac{Base}{Acid} = 0.5

    \frac{Base}{Acid} = 3.162

Now,

        \frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

                 

8 0
4 years ago
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