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Crank
3 years ago
8

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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Which of the following is not true? a. The freezing point of sea water is lower than the freezing point of pure water. b. The bo
Anna007 [38]

Answer:

The false statement is b.

The boiling point of a water sample from the Salt Lake is lower than the boiling point of pure water

Explanation:

This excersise refers to colligative properties.

Boiling point refers to the property of boiling point elevation. In a solution of ions, as water sea or water from lakes, the boiling point will be higher than water pure.

It is logical to say that, because these sort of water have ions. Remember that colligative properties depends on the solute particles.

T° boiling solution - T° pure solvent = Kb . m . i

In the freezing point depression, we have the oppossite of boiling point elevation. Freezing point of solution is lower than pure solvent, according to this:

T° freezing pure solvent - T° freezing solution = Kc . m . i

We have to always consider the i, which means Van't Hoff factor, number of ions dissolved in solution. As the i is higher, the freezing point of solution will be lower, and the boiling point of solution will be higher.

They are true statement a and c.

In solution of., [NaCl] = 0.1M, as it is a higher concentration, the molality is also higher than a solution of [NaCl] = 0.05.

As water sample form Salt Lake has certain ions, it is logical to say that the boiling point of this water is higher than boiling point of pure water.

5 0
3 years ago
What are the balanced chemical equation for these acid and base reactions?
photoshop1234 [79]
It can be easily judged that only the first reaction is an acid base reaction among the three given in the question. So, we can avoid the other two reactions given in the question. Now let us focus and write down the balanced chemical equation of <span>P-Toluidine + HCl.

</span><span>C7H9N + HCl = C7H10N (+) + Cl (-)
</span>
I hope the answer has come to your help.
6 0
3 years ago
A rigid plastic container holds 1.00 l methane gas at 660 torr pressure when the temperature is 22.0 degrees Celsius. How much p
igor_vitrenko [27]
 <span>PV/T = P'V'/T' 
660 x 1.00/295.2 = P' x 1.00/317.8 
P'=710.5 torr</span>
6 0
3 years ago
Read 2 more answers
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
Ozone has a molecular formula of O3. If 7.92*1024 atoms of oxygen react to form ozone, how many
prisoha [69]

337.92 moles of Ozone will be produced

1 Oxygen atom is 8 g

1 mole of ozone, O3 = 8 * 3 = 24 g

7.92 * 1024 = 8110.08 g

1 mole = 24 g

? moles = 8110.08 g

? = 337.92 moles


Read more on moles here:

brainly.com/question/15356425

Hope it helps

7 0
2 years ago
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