Last post as a troll lol, sorry but i need actual help this time.

How many formula units make up 24.2 g of magnesium chloride (MgCl2)? 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 $0.254\; \rm mol$ :

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

6 0

3 years ago

If a gas is initially at a pressure of nine ATM and a volume at 21 L at a temperature of 253K and the pressure is raise to 15 AT

Tom [10]

Answer:

15.0L

Explanation:

p/v = constan

(9*21)/253 =(15v)/ 302

v = (9*21*302)/(15*253)

v=15.0

3 0

3 years ago

In just thomsons experiments with electricity he showed that an electrical current can be

katen-ka-za [31]

In Thomson's experiment, he showed that an electrical current can be made to flow from a positive site to a negative site.

6 0

3 years ago

Read 2 more answers

A mole is the amount of a substance that contains as many particles as the number of atoms in 12 grams of what isotope?

Ray Of Light [21]

The SI unit for amount of substance is the mole. It has the unit symbol mol. The proportionality constant is the inverse of the Avogadro constant. The mole is defined as the amount of substance that contains an equal number of elementary entities as there are atoms in 12g of the isotope carbon-12.

Hope This Helped! :3

3 0

3 years ago

how many ml of a 22.5% (v/v) ethanol solution would you need to measure out in order to have 12.5 ml of ethanol ?

Aleks [24]

Volume percent or volume/volume percent (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:

12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed

Hope this answers the question.

8 0

4 years ago