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Elza [17]
3 years ago
12

Element 2 is: A) Cobalt B)Chlorine C)Calcium D)Carbon

Chemistry
1 answer:
saul85 [17]3 years ago
4 0

Answer:

D: Carbon

Explanation:

Carbon is the sixth element with a total of 6 electrons in the periodic table. Hence the atomic number Z = 6. The ground state electron configuration of carbon is 1s2 2s2 2p2. An excited state electron configuration of carbon is 1s2 2s1 2p3.

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How many moles of hydrogen are needed to produce 13.78 mol of ethane?
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Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.
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What other traits besides phisical ones could be passed on from parent offspring​
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Answer:

Love for Music

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2 years ago
Which is a living part of an ecosystem?<br><br>A.soil<br>B.bacteria<br>C.animal bones<br>D.minerals​
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Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)
AURORKA [14]

Explanation:

The net equation will be as follows.

          K(s) + Cl_{2}(g) \rightarrow KCl(s)

So, we are required to find \Delta H_{formation} for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1:  Convert K from solid state to gaseous state

          K(s) \rightarrow K(g),    \Delta H_{1} = 89 kJ

Step 2:  Ionization of gaseous K

           K(g) \rightarrow K^{+}(g) + e^{-},    H_{2} = 418 KJ

Step 3:  Dissociation of Cl_{2} gas into chlorine atom .

            \frac{1}{2} Cl_{2}(g) \rightarrow Cl(g),   \Delta H_{3} = \frac{244}{2} = 122 KJ

Step 4: Iozination of chlorine atom.

              Cl(g) + e^{-} \rightarro Cl^{-}(g),      H_{4} = -349 KJ

Step 5:  Add K^{+} ion and Cl^{-} ion formed above to get KCl .

              K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s),   H_{5} = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

      \Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}

                  = 89 + 418 + 122 - 349 - 717

                  = - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

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Which of these CANNOT happen in a chemical reaction ?
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The production of mew atoms?
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