Tap water used for drinking is considered in the DIL calculations. A. True B. False

What is the formula for carbon tetrachloride? a. CCl b. CCl2 c. CCl3 d. CCl4

Fittoniya [83]

The formula for carbon tetrachloride is CCl4. The answer is letetr D. The rest of the choices do not answer the question above.

7 0

3 years ago

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

2 years ago

100 POINTS! Final Honor Activity Question

castortr0y [4]

The change in temperature had the greatest effect at changing the volume of the balloon.

The gas laws are used to describe the parameters that has to do with gases.

Given that;

P1 = 98.5 kPa

T1 = 18oC or 291 K

V1 = 74.0 dm3

P2 = 7.0 kPa

V2 = ?

T2 = 18oC or 291 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V2= P1V1T2/P2T1

V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K

V2 = 1041.3 dm3

When;

V1 = 1041.3 dm3

T1 = 291 K

V2 = ?

T2 = 80oC or 353 K

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2/T1

V2 = 1041.3 dm3 * 353 K/291 K

V2 = 1263 dm3

The change in temperature had the greatest effect at changing the volume of the balloon.

Given that

V1 = 100 cm^3

T1 = 273 K

P1 = 1.01 * 10^5 Pa

V2 = ?

P2 = 3.00 x 10^-4 Pa

T2 = -180oC or 255 K

V2= P1V1T2/P2T1

V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K

V2 = 3.14 * 10^10 cm^3

Learn more about gas laws:brainly.com/question/12669509

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7 0

2 years ago

Estimate the air pressure at an altitude of 5km

r-ruslan [8.4K]

The answer is 100 kg

6 0

3 years ago

I need help with this anyone?

valkas [14]

Just look it up on goog^le or a chart

5 0

3 years ago