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balandron [24]
3 years ago
9

Which of the following does not form a diatomic molecule?

Chemistry
2 answers:
Sunny_sXe [5.5K]3 years ago
6 0
For the answer to the question above, Copper does not form a diatomic molecule. A diatomic molecule has only two atoms. Diatomic molecules are the molecule that is composed of two same atoms which are bonded covalently. An example is an oxygen, it exists as O2 and chlorine, it exists as Cl2. I hope my answer helped you.
boyakko [2]3 years ago
3 0
A diatomic molecule is one that forms a bond between two of the same molecule. The main diatomic molecules are Hydrogen, Oxygen, Nitrogen, Fluorine, Chlorine, Bromine, and Iodine. There are many reasons in quantum mechanics for this answer, but the simplest way to think about this is that the atoms want to have an outer shell of eight electrons, except for hydrogen, which wants to have two. Hydrogen, nitrogen, and chlorine can all bond with another atom of the same type to satisfy their full outer shell. Copper, on the other hand, does not form with one other copper to form Cu2, so the answer is D. Copper.
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3 years ago
How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!
NNADVOKAT [17]

Answer:

Approximately 1.53 \times 10^{23} formula units (0.254\; \rm mol).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium (\rm Mg) and chlorine (\rm Cl):

  • \rm Mg: 24.305.
  • \rm Cl: 35.45.

In other words, the mass of 1\; \rm mol of \rm Mg atoms would be (approximately) 24.305\; \rm g.

Likewise, the mass of 1\; \rm mol of \rm Cl atoms would be approximately 35.45\; \rm g.

One formula unit of the ionic compound \rm MgCl_{2} includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of 1\; \rm mol of the formula units of this compound.

The formula \rm MgCl_{2} includes one \rm Mg atom and two \rm Cl atoms.

Hence, every formula unit of \rm MgCl_{2} \! would include the same number of atoms: one \rm Mg\! atom and two \rm Cl\! atoms. There would be 1\; \rm mol of \rm Mg atoms and 2\; \rm mol of \rm Cl atoms in 1\; \rm mol\! of \rm MgCl_{2} formula units.

Thus, the mass of 1\; \rm mol\! of \rm MgCl_{2} formula units would be equal to the mass of 1\; \rm mol of \rm Mg atoms plus the mass of 2\; \rm mol of \rm Cl atoms. (The mass of 1\; \rm mol\!\! of each atom could be found from the relative atomic mass of each element.)

\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}.

In other words, the formula mass of \rm MgCl_{2} is 95.205\; \rm g \cdot mol^{-1}.

Therefore, the number of formula units in m = 24.2\; \rm g of \rm MgCl_{2} would be:

\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}.

Multiple n by Avogadro's Number N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1} to estimate the number of formula units in 0.254\; \rm mol:

\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}.

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A student calculated the density of a sample of graphite to be 2.3 g/cm3. Show a numerical setup for calculating the student’s p
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Answer:

Percent error = 1.5%

Explanation:

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Now we will put the values:

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Answer:

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