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3 years ago

What is the molecular formula for chloroform? none of these choices CH3C CHCI3 CH2C12 CCI4

Chemistry

1 answer:

dangina [55]3 years ago

4 0

Answer:

CHCI₃

Explanation:

Chloroform, IUPAC name, trichloromethane, is organic compound with the molecular formula CHCl₃. It is colorless and sweet-smelling liquid having high density which is produced on a large scale precursor of PTFE , and for various refrigerants .

Chloroform , is a powerful euphoriant , anxiolytic , anesthetic and sedative when inhaled or ingested .

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Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

3 years ago

What is Lewis acid and Lewis base? give examples

AleksAgata [21]

Explanation:

example is copper iron...........

6 0

3 years ago

What is the atomic number of the element located in group 16

laila [671]

Answer:

Sulfur

Explanation:

Sulfur is a chemical element that is represented with the chemical symbol "S" and the atomic number 16 on the periodic table.

7 0

3 years ago

To gravimetrically analyze the silver content of a piece of jewelry made from an alloy of Ag and Cu, a student dissolves a small

Ipatiy [6.2K]

Answer:

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions.

Explanation:

The student have in solution Ag⁺ and Cu²⁺ ions but he just want to analyze the silver, that means he need to separate ions.

Centrifuging the solution to isolate the heavier ions FALSE Centrifugation allows the separation of a suspension but Ag⁺ and Cu²⁺ are both soluble in water.

Adding enough base solution to bring the pH up to 7.0 FALSE At pH = 7,0 these ions are soluble in water and its separation will not be possible.

Adding a solution containing an anion that forms an insoluble salt with only one of the metal ions TRUE For example, the addition of Cl⁻ will precipitate the Ag⁺ as AgCl(s) allowing its separation.

Evaporating the solution to recover the dissolved nitrates. FALSE . Thus, you will obtain the nitrates of these ions but will be mixed doing impossible its separation.

I hope it helps!

8 0

3 years ago

What is the final volume (L) of a 1.00 L system at 315 K and 1.10 atm if STP conditions are established?

LenaWriter [7]

Answer:

0.95L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 1L

T1 (initial temperature) = 315K

P1 (initial pressure) = 1.10 atm

T2 (final temperature) = stp = 273K

P2 (final pressure) = stp = 1atm

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the system can be obtained as follow:

P1V1/T1 = P2V2/T2

1.1 x 1/315 = 1 x V2/273

Cross multiply to express in linear form.

315 x V2 = 1.1 x 273

Divide both side by 315

V2 = (1.1 x 273) /315

V2 = 0.95L

Therefore, the final volume of the system if STP conditions are established is 0.95L

5 0

3 years ago