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snow_tiger [21]
2 years ago
9

Titanium dioxide, TiO₂, reacts with carbon and chlorine to give gaseous TiCl₄: TiO₂+2C+2CI₂−TiCI₄+2CO The reaction of 7.39 kg ti

tanium dioxide with excess C and Cl₂ gives 14.24 kg titanium tetrachloride. Calculate the theoretical yield of TiCl₄ (assuming complete reaction) and its percentage yield.
Chemistry
1 answer:
Dima020 [189]2 years ago
3 0

Answer:

17.57kg of TiCl_{4} and its percentage yield is 81.0%

Explanation:

Through the reaction you can get the theoretical amount of  TiCl_{4} that must be produced.

7.39kgTiO_{2}x\frac{1kmolTiO_{2} }{79.867kgTiO_{2}}x \frac{1kmolTiCl_{4}}{1kmolTiO_{2}}x\frac{189.867kgTiCl_{4} }{1kmolTiCl_{4}}=17.57kgTiCl_{4}

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage

yield=\frac{actual amount}{theoretical amount}x100= \frac{14.24kg}{17.57kg}x100=81.0%

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Why are familiar objects such as pens and paper clips not commonly counted in moles?
Anton [14]
Well ask yourself why don't we count it in moles and you should get your answer.
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Large quantities of tritium were released into the atmosphere in the 1950s and 1960s during atmospheric nuclear testing. This tr
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What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solut
kherson [118]

Answer:

P = 17.9618 atm

Explanation:

The osmotic pressure can be calculated and treated as if we are talking about an ideal gas, and it's expression is the same:

pV = nRT

However the difference, is that instead of using moles, it use concentration so:

p = nRT/V ----> but M = n/V so

p = MRT

We have the temperature of 18 °C (K = 18+273.15 = 291.15 K) the value of R = 0.08206 L atm / K mol, so we need to calculate the concentration, and we have the mass of HCl, so we use the molar mass of HCl which is 36.45 g/mol:

n = 13.7/36.45 = 0.3759 moles

M = 0.3759/0.5 = 0.7518 M

Now that we have the concentration, let's solve for the osmotic pressure:

p = 0.7518 * 0.08206 * 291.15

<em><u>p = 17.9618 atm</u></em>

3 0
2 years ago
The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C
Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 460^oC = 5.8\times 10^{-6}s^{-1}

K_2 = rate constant at T_2 = 4\times K_1

Ea = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 460^oC=273+460=733K

T_2 = final temperature = ?

Now put all the given values in this formula, we get:

\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]

T_2=791.1K

Therefore, the final temperature would be, 791.1 K

4 0
2 years ago
Express each aqueous concentration in the unit indicated.
MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L =  9500 millileters (ml)

we consider 25 mg = 25 millileters

<em>ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)</em>

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml =  10000  × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

<em>ppm = (mass of solute / mass of solvent) × 1000000 (1 million)</em>

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

<em>moles = mass / molar mass</em>

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

<em>Molarity = moles of solute / litres of solution</em>

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0
2 years ago
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