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Tems11 [23]
3 years ago
11

Combustion analysis of toluene, a common organic solvent, gives 3.52 mg of co2 and 0.822 mg of h2o. if the compound contains onl

y carbon and hydrogen, what is its empirical formula?
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0
<span>C7H8 First, lookup the atomic weight of all involved elements Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Then calculate the molar masses of CO2 and H2O Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now calculate the number of moles of each product obtained Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK. moles CO2 = 3.52 / 44.0087 = 0.079984 moles H2O = 0.822 / 18.01488 = 0.045629 Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen. moles C = 0.079984 moles H = 0.045629 * 2 = 0.091258 So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting 1: 1.140953 The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like) 1 ( 1 : 1.140953) = 1 : 1.140953 2 ( 1 : 1.140953) = 2 : 2.281906 3 ( 1 : 1.140953) = 3 : 3.422859 4 ( 1 : 1.140953) = 4 : 4.563812 5 ( 1 : 1.140953) = 5 : 5.704765 6 ( 1 : 1.140953) = 6 : 6.845718 7 ( 1 : 1.140953) = 7 : 7.986671 8 ( 1 : 1.140953) = 8 : 9.127624 That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
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lina2011 [118]
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8 0
3 years ago
While you were "sweating" your chemistry test, water vapor evaporates from your body, absorbing 50,000 J of energy. (assume no
TiliK225 [7]

Answer:

\boxed {\boxed {\sf B. \ 22 \ grams}}

Explanation:

We need to use the formula for heat of vaporization.

Q=H_{vap}*m

Identify the variables.

  • The heat absorbed by the evaporating water is the <u>latent heat of vaporization. </u>For water, that is 2260 Joules per gram.
  • Q is the energy, in this problem, 50,000 Joules.
  • m is the mass, which is unknown.

H_{vap}=2260 \ J/g\\Q=50,000 \ J \\

Substitute the values into the formula.

50,000 \ J=2260 \ J/g*m

We want to find the mass. We must isolate the variable, m.

m is being multiplied by 2260 J/g. The inverse operation of multiplication is division. Divide both sides by 2260 J/g.

\frac{50,000 \ J}{2260 \ J/g} =\frac{2260 \ J/g*m}{2260 \ J/g}

\frac{50,000 \ J}{2260 \ J/g} =m

Divide. Note that the Joules (J) will cancel each other out.

\frac{50,000 \ }{2260 \ g} =m

22.1238938 \ g =m

Round to the nearest whole number. The 1 in the tenth place tells us to leave the number as is.

22 \ g \approx m

The mass is about 22 grams, so choice B is correct.

3 0
3 years ago
What is the mass of the block of iron illustrated below?
Mandarinka [93]
VOLUME= 5cm*10cm*2cm =100cm^3
but density of iron=7.874g/cm^3
mass=7.874g*100 =787.4g
mass of that block = 787.4g
7 0
3 years ago
Read 2 more answers
A particular sample of pure iron requires 0.612 kJ of energy to raise its temperature from 30.°C to 51°C. What must be the mass
mixas84 [53]

Answer:

m = 65.637 g

Explanation:

Heat = 0.612 kJ = 612 J ( Converting to J by multiplying by 1000)

Initial Temperature =  30.°C

Final Temperature = 51°C

Temperature change = Final Temperature - Initial Temperature = 51 - 30 = 21°C

Mass = ?

The relationship between these quantities is given by the equation;

H = mCΔT

where c = 0.444 J/g°C

Inserting the values in the equation;

612 = m *  0.444 * 21

m = 612 / (0.444 * 21)

m = 65.637 g

3 0
2 years ago
Help me out with these three formulas
ddd [48]
Carbon dioxide it should be water oxygen
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