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emmainna [20.7K]
3 years ago
6

Im doing a science project and need examples and non-examples of an Atom. some examples of an atom is neon, hydrogen, argon, etc

so I got that part. But what are some non-examples of an atom and why?​
Chemistry
1 answer:
gogolik [260]3 years ago
6 0

Answer:

Anything not on the periodic table is an element non example! ... So, for a substance to be an element, all of its atoms must have the same number of protons. Examples of elements include hydrogen, lithium, nickel, and radium.

Explanation:

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What impact does claim ate have on biodiversity​
SashulF [63]

Do you mean climate? If so;

Essentials, healthy ecosystems and rich biodiversity are fundamental to life on our planet. Climate is affecting the habitats of several creatures, which must migrate to areas with better conditions. Even small changes in average temperatures can have a significant effect upon ecosystems.

3 0
3 years ago
Consider the following chemical reaction.
Ede4ka [16]

The concentration of the zinc acetate, Zn(C₂H₃O₂) solution is 1.067 M

  • We'll begin by calculating the number of mole in 20.5 g of Zn(C₂H₃O₂)₂

Mass of Zn(C₂H₃O₂)₂ = 20.5 g

Molar mass of Zn(C₂H₃O₂)₂ = 65 + 2[(12×2) + (3×1) + (16×2)]

= 65 + 2[59]

= 183 g/mol

<h3>Mole of Zn(C₂H₃O₂)₂ =? </h3>

Mole = mass / molar mass

Mole of Zn(C₂H₃O₂)₂ = 20.5 / 183

<h3>Mole of Zn(C₂H₃O₂)₂ = 0.112 mole </h3>

  • Finally, we shall determine concentration of Zn(C₂H₃O₂). This can be obtained as follow:

Mole of Zn(C₂H₃O₂)₂ = 0.112 mole

Volume = 105 mL = 105 / 1000 = 0.105 L

<h3>Concentration of Zn(C₂H₃O₂)₂ =? </h3>

Concentration = mole / Volume

Concentration of Zn(C₂H₃O₂)₂ = 0.112 / 0.105

<h3>Concentration of Zn(C₂H₃O₂)₂ = 1.067 M</h3>

Therefore, the concentration of the zinc acetate, Zn(C₂H₃O₂) solution is 1.067 M

Learn more: brainly.com/question/25511880

5 0
2 years ago
Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration
gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}

        K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}

Let us assume that the solubility be "s". And, the reaction equation is as follows.

        AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

     9.84 \times 10^{-21} = s \times s

             s = 9.92 \times 10^{-11}

Also,     Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}

                2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}

                            s = 7.14 \times 10^{-7}

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the K_{sp} expression as follows.

         K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}

          2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}

       2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}

       [PO^{3-}_{4}]^{2} = 4.88 \times 10^{-24}

                             = 2.21 \times 10^{-12} M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

             AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

           K_{sp} = [Al^{3+}][PO^{3-}_{4}]

       9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}

                [Al^{3+}] = 4.45 \times 10^{-9} M

Thus, we can conclude that concentration of aluminium will be 4.45 \times 10^{-9} M when calcium begins to precipitate.

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Answer:

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