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lubasha [3.4K]
3 years ago
6

How many total moles of ions are released when the following sample dissolves completely in water? 6.188 x 10 21 formula units o

f NiCl 2 a. 1.36 mol of ions b. 1.86 mol of ions c. 3.08 x 1022 mol of ions d. 0.0205 mol of ions e. 3.08 x 10-2 mol of ions
Chemistry
1 answer:
elixir [45]3 years ago
3 0

Answer:

e. 3.08 x 10⁻² mol of ions.

Explanation:

  • Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).

  • We can get the no. of moles of NiCl₂ using cross multiplication:

1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.

??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.

∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.

  • NiCl₂ is ionized according to the equation:

NiCl₂ → Ni²⁺ + 2Cl⁻.

Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).

<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>

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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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Answer:

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How many atoms are in 578g Na
borishaifa [10]

1.51 x 10²⁵atoms

Explanation:

Given parameters:

Mass of Na = 578g

Unknown:

Number of atoms = ?

Solution:

 To find the number of atoms, we must first find the number of moles the given mass contains.

  Number of moles  = \frac{mass}{molar mass}

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 This gives  1.51 x 10²⁵atoms of Na

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