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exis [7]
3 years ago
12

How many moles is 130g of CoCl2

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

1 mol

Explanation:

The mass of CoCl2 is 129.83. 130 divided by the mass is 1. This is to convert the mass given to mols that exist in the sample.

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to have the closest number rounded up

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Draw a picture of what you imagine solid sodium chloride looks like at the atomic level. (Do NOT draw Lewis structures.) Make su
GenaCL600 [577]

Answer:

Kindly check the explanation section.

Explanation:

PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.

The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and

Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.

In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.

7 0
3 years ago
Suggest a name for a compound containing potassium, carbon and oxygen.
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Answer:

Oxidation of potassium amalgam with carbon dioxide results in the formation of potassium oxalate. Potassium is not reactive with benzene, although heavier alkali metals such as cesium react to give organometallic products.

Please Mark Brainliest If This Helped!

4 0
2 years ago
How are acids and bases used by humans
azamat
We use acids and bases in everyday life for cooking and cleaning
6 0
3 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.

7 0
3 years ago
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