A 0.500 L bottle of water contains 8.82 × 10–7 mL of benzene. What is the concentration of benzene in ppmv?

For the reaction below, Kp 5 1.16 at 800.8C. CaCO3(s) 34 CaO(s) 1 CO2(g) If a 20.0-g sample of CaCO3 is put into a 10.0-L contai

Elena L [17]

Answer:

The mass percentage of calcium carbonated reacted is 2.5%.

Explanation:

The reaction is:

$CaCO_{3}(s)--->CaO(s)+CO_{2}(g)$

Thus the Kp of the equilibrium will be:

Kp = partial pressure of carbon dioxide [as the other are solid]

Moles of calcium carbonate initially present = $\frac{mass}{molarmass}=\frac{20}{100}=0.2$

Let us apply ICE table to the equilibrium given:

$CaCO_{3}(s)--->CaO(s)+CO_{2}(g)$

Initial 0.2 0 0

Change -x +x +x

Equilibrium 0.2-x x x

Kp = partial pressure of carbon dioxide

Kp = Kc(RT)ⁿ

where n = difference in the number of moles of gaseous products and reactants

for given reaction n = 1

R = gas constant = 8.314 J /mol K

T = temperature = 800 ⁰C = 1073 K

Putting values

Kc = $\frac{Kp}{RT}=\frac{1.16}{8.314X1073}=1.3X10^{-4}$

Kc = $\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}$

$1.3X10^{-4}(0.2-x)=x^{2}$

$x^{2} = 0.26X10^{-4}-1.3X10^{-4}x$

On calculating

x = 0.005

where x = the moles of calcium carbonate dissociated or reacted.

Percentage of the moles or mass reacted = $\frac{molesreacted X100}{initialmoles}=\frac{0.005X100}{0.2}=2.5$ %

7 0

2 years ago

How many moles of PBr3 contain 3.68 x 10^25 bromine atoms?

scoray [572]

3.68 x 10²⁵ bromine atoms * 1mol/6.02*10²³ atoms=
= 61.13 mol of bromine atoms

1 mol PBr3 ----- 3 mol Br
x mol PBr3 -----61.13 mol Br

x= 1*61.13/3 = 20.4 mol PBr3.

20.4 mol PBr3 contain 3.68 x 10^25 bromine atoms.

7 0

2 years ago

Read 2 more answers

Suppose that magnesium would react exactly the same as copper in this experiment. how many grams of magnesium would have been us

Vikentia [17]

The solution for this problem would be:

We are looking for the grams of magnesium that would have been used in the reaction if one gram of silver were created. The computation would be:

1 g Ag (1 mol Mg) (24.31 g/mol) / (2mol Ag)(107.87g/mol) = 0.1127 grams of Magnesium

6 0

2 years ago

When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po

SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide ( $C_{7}H_{7}NO$ ) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is $2.7^{o}C$ lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride ( $NH_{4}Cl$ ) are dissolved in the same mass of X, the freezing point of the solution is $9.9^{o}C$ lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

Moles of benzamide = $\frac{mass}{\text{Molar mass of benzamide}}$

= $\frac{70.4 g}{121.14 g/mol}$

= 0.58 mol

Now, we will calculate the molality as follows.

Molality = $\frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}$

= $\frac{0.58 mol}{0.85 kg}$

= 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

dT = $i \times K_{f} \times m$ ,

where, dT = change in freezing point = $2.7^{o}C$

i = van't Hoff factor = 1 for non dissociable solutes

$K_{f}$ = freezing point constant of solvent

m = 0.6837

Therefore, putting the given values into the above formula as follows.

dT = $i \times K_{f} \times m$ ,

$2.7^{o}C = 1 \times K_{f} \times 0.6837 m$

$K_{f}$ = 3.949 C/m

Now, we use this $K_{f}$ value for calculating i for $NH_{4}Cl$

So, moles of ammonium chloride are calculated as follows.

Moles of $NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}$

= 1.316 mol

Hence, calculate the molality as follows.

Molality = $\frac{1.316 mol}{0.85 kg}$

= 1.5484

It is given that value of change in temperature (dT) = $9.9^{o}C$ . Thus, calculate the value of Van't Hoff factor as follows.

dT = $i \times K_{f} \times m$

$9.9^{o}C = i \times 3.949 C/m \times 1.5484 m$

i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0

2 years ago

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1

erastovalidia [21]

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

c. 81% by mass CH₃CN

Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, in this case / Total mass) . 100

We were asked for the acetonitrile → (667.7 g / 824 g) . 100 = 81%

3 0

2 years ago