All categories

3 years ago

How many particles are present in 15.0 g of CCL4? (Show work please:))

Chemistry

2 answers:

Fed [463]3 years ago

7 0

Answer:

5.86*10^22particles

Explanation:

First obtain the molar mass of tetrachloromethane then compare the given mass in the question with Avogadro's number as shown in the solution. Note that the Avogadro's number is the number of particles contained by one mole of all substances at standard temperature and pressure. According to Avogadro, at the same temperature and pressure, one mole of all substances contain equal number of molecules

SCORPION-xisa [38]3 years ago

6 0

1) Calculate the molar weight. 12.000 + 4*(35.45)
2) Divide 15.0 by the molar weight to get moles
3) Multiply by Avogado's number of get the number of molecules.

You might be interested in

What will happen when the rates of evaporation and condensation are equal

Ilia_Sergeevich [38]

A random person put the answer for you

6 0

3 years ago

Why do we call water the univeral salivnet

ankoles [38]

Water<span> is capable of dissolving a variety of different substances, which is why it is such a good </span>solvent. Water dissolves more than any other liquid, therefor we called it the Universal Solvent.

6 0

2 years ago

Stock naming please help

andre [41]

Answer:

Can you tell me the question in the comments on this answer or like how you do it then ill answer you in the comments under this answer

8 0

2 years ago

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$

$pOH=-log[OH^-}$

$pH+pOH=14$

$[H_3O^+][OH^-]=10^{-14}$

a. Solution A: $[OH^-]=3.33\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M$

b. Solution B : $[H_3O^+]=9.33\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M$

c. Solution C : $[H_3O^+]=5.65\times 10^{-4}M$

$[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M$

7 0

2 years ago

Can u please help me I don't understand it and if I don't do it my teacher will get really mad at me pleas

Nana76 [90]

Physical Change: It is a type of change in which matter changes its physical state like shape, size but is not transformed into another substance. It is usually a reversible process.

Chemical Change: It is a type of change in which the rearrangement of atoms of one or more than one substance is involved. and it changes its chemical composition that is there is a formation of at least one new substance. It is usually an irreversible process.

Now, keeping in mind the definitions, we can easily classify the examples in the question as physical or chemical change.

7. Chemical Change

8. Chemical Change

9. Physical Change

10. Chemical Change

11. Physical Change

12. Physical Change

13. Chemical Change

14. Physical Change

15. Chemical Change

16. Physical Change

17. Chemical Change

18. Chemical Change

19. Physical Change

20. Physical Change

21. Chemical Change

22. Physical Change

23. Chemical Change

24. Chemical Change

25. Physical Change

6 0

3 years ago