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igomit [66]
3 years ago
13

Iron has a density of 7.86 g/cm3. Calculate the volume (in dL) of a piece of iron having a mass of 3.11 kg

Chemistry
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:

The volume is 3.96 dL

Explanation:

<u>Step 1: Convert ml to dl</u>

1Liter = 1000ml = 10³ ml

1Liter = 10 dl

⇒ 1 Liter has 1000ml or 10 dl

⇒ The ratio can be written as : 10dl/ 1000ml = 1 dl / 100ml

<u>Step 2: Convert cm³ </u>=

⇒ 1cm³ is equal to 1 ml.

<u>Step 3:  Calculate volume</u>

⇒ We use the formule of density here, which is : Density (p) = mass (g) / volume( cm³)

⇒ 7.86 g/ cm³  = 3110 grams / Volume ( cm³)

⇒Volume = 3110 grams / 7.86g/cm³ = 395.67 cm³

395.67 cm³ = 395.67 ml = 0.39567 L  = 3.96 dL

3.96 dL is reported to three significant digits due to the precision reported for the density.

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There are 1.93 x 10²⁴ particles

<h3>Further explanation</h3>

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\tt 3.2~moles\times \dfrac{6.02\times 10^{23}}{1~mole}=1.93\times 10^{24}

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How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!
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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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