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3 years ago

A 258-ml sample of mek, an industrial solvent, is found to have a mass of 0.599 g at and 100c. what is the molar mass of mek?

Chemistry

2 answers:

Kobotan [32]3 years ago

5 0

Answer:

The ans is 71.1 g/mol

Explanation:

Mass of MEK (m) = 0.599 g

Volume of MEK (V) = 258 ml = 0.258 L

Temperature (T) = 100 C = 100 +273 = 373 K

Pressure = 1 atm

Based on the ideal gas equation:

$PV = nRT$

where n = number of moles and R = gas constant 0.0821 Latm/mol-K

$n = \frac{PV}{RT}$

$n = \frac{1*0.258}{0.0821*373} = 0.00842 moles$

number of moles is given as:

$n = \frac{mass MEK}{molar mass MEK} \\\\molar mass = \frac{mass MEK}{n} = \frac{0.599}{0.00842} = 71.14 g/mol$

SVETLANKA909090 [29]3 years ago

4 0

Correct answer is : 72.0 g/mole If we assume Ideal Gas Law, PV = nRT. n = PV/RT = ((0.988 atm)*(0.258 L))/((0.0821 L*atm/mole*K)*(100 + 273.15)) = 0.00832 moles Molar mass = 0.599 g/0.00832 moles = 72.0 g/mole; your answer is E.

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A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is pr

s344n2d4d5 [400]

A solution is prepared by adding 1.43 mol of potassium chloride (kcl) to 889 g of water. The concentration of kcl is 1.61 molal.
mol of Kcl (potassium chloride)= 1.43
water = 889 g
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3 years ago

The atomic number of fluorine is 9. How many electrons does an ion of fluorine have if it is represented by the symbol shown bel

Irina-Kira [14]

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Look at the EC

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2 years ago

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Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

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

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

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3 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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2 years ago