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timurjin [86]
3 years ago
6

1.00 g of a compound is combusted in oxygen and found to give 3.14g of CO2 and 1.29 g of H2O. From these data we can tell thatA.

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula.B. the compound contains only C and H and has the empirical formulaof C6H.C. the compound contains C, H, and O and has the empirical formulaof CH3O.D. the compound contains only C and H and has the empirical formulaof CH2.E. None of the above is a true statement
Chemistry
1 answer:
lora16 [44]3 years ago
5 0

Answer:

the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula

Explanation:

Mass of CO2 obtained = 3.14 g

Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol

The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g

Mass of H2O obtained = 1.29 g

Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol

The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g

% by mass of carbon = 0.857/1 ×100 = 85.7 %

% by mass of hydrogen = 0.0717/1 × 100 = 7.17%

Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %

Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.

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A solution is prepared by dissolving 15 grams of sugar in 85 grams of water. How many percent of sugar is in the solution?
trasher [3.6K]

Answer:

15%

Explanation:

From the question given above, the following data were obtained:

Mass of sugar = 15 g

Mass of water = 85 g

Percentage of sugar in the solution =?

Next, we shall determine the mass of the solution. This can be obtained as follow:

Mass of solute (sugar) = 15 g

Mass of solvent (water) = 85 g

Mass of solution =?

Mass of solution = mass of solute + mass of solvent

Mass of solution = 15 + 85

Mass of solution = 100 g

Finally, we shall determine the percentage of the sugar in the solution. This can be obtained as follow:

Mass of solute (sugar) = 15 g

Mass of solution = 100 g

Percentage of sugar in the solution =?

Percentage = solute /solution × 100

Percentage = 15 / 100 × 100

Percentage of sugar in the solution = 15%

7 0
2 years ago
What volume is occupied by 22.1 g of O2 at 52°C and a pressure of 1.63 atm?
Nutka1998 [239]
U can use PV=nRT
where n=mass in grams/ molar mass
temp will be converted into kelvin by adding 273k
the value of R = 0.0821
try it by yourself :)
3 0
3 years ago
A current is applied to two electrolytic cells in series. In the first, silver is deposited; in the second, a zinc electrode is
vitfil [10]

The amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Given ,

A current is applied to two electrolytic cells in series .

In the first ,silver is deposited and in the second a zinc electrode is consumed .

the reactions involving are ;

Ag+ (aq) + e = Ag

Zn = Zn2+ (aq) +2e

thus the resultant equation is ,

2Ag+ (aq) +Zn = 2Ag + Zn2+

Thus for every mole of Zn dissolves , there is 2 moles of Ag is formed .

65.38 g of Zn contains = 1 moles

1.2 g of Zn contains = 1.2/65.38 =0.01835 moles

for every 1 mole of Zn dissolves there is 2 moles of Ag formed .

Thus the amount of Ag formed in moles =2(O.01835) =0.0367 Moles

1 mole of Ag contains = 107.86g

0.0367 moles of Ag contains = 107.86 (0.0367) =3.959 g of Ag

Hence ,the amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Learn more about electrolytic cell here:

brainly.com/question/19854746

#SPJ4

6 0
1 year ago
Why is the time series useful as a management tool?
katrin [286]

Answer:

A)

Explanation:

It estimates the probability of extinction.

8 0
3 years ago
The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C
artcher [175]

Explanation:

The given data is as follows.

         P_{1} = 100 mm Hg or \frac{100}{760}atm = 0.13157 atm

         T_{1} = 1080 ^{o}C = (1080 + 273) K = 1357 K

         T_{2} = 1220 ^{o}C = (1220 + 273) K = 1493 K

         P_{2} = 600 mm Hg or \frac{600}{760}atm = 0.7895 atm

          R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

                   log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}

            log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]

          log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]

                0.77815 = \frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K

              \Delta H_{vap} = 2.219 \times 10^{5} J/mol

                                   = 2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}

                                    = 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0
2 years ago
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