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laiz [17]
3 years ago
8

What are you and what do you want with my attendance

Chemistry
1 answer:
Lelu [443]3 years ago
8 0

Answer:

Such a waste of points but lol

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How many atoms are in 1.75 mol CHCL3
vaieri [72.5K]

1 mol = 6.022 x 10²³ atoms

In order to find how many atoms, dimly multiply the amount of moles you have by 6.022 x 10²³ or Avogadro's number.

So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃

But now you have to round because of the rules of significant figures so you get 1.05 x 10²⁴ atoms of CHCl₃

6 0
3 years ago
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PLEASE EXPLAIN CORRECTLY! THX=)
tankabanditka [31]
A way to explain it is that back then all the continents were together but soon after drifted apart the were in the same place sort of put they drifted apart so that's sorta what happened. Hope that helps a little
8 0
3 years ago
Which of the following reasons best explains why water isn
Sholpan [36]

Answer:

the wax is a nonpolar substance that will not mix with polar water

Explanation:

Water is polar by due to the uneven distribution of charge between the hydrogen and oxygen. T=This automatically eliminates options 3 and 4.

Wax on the other hand is a non polar substance. Due to this non polar characteristic, it would not dissolve in water. The correct option is;

- the wax is a nonpolar substance that will not mix with polar water

3 0
3 years ago
When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

m= molality = \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579

8.30^0C=1\times K_f\times 0.579

K_f=14.3^0C/m

Let Mass of solute (KBr) = x g

8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0
2 years ago
Helpp me pretty please
lubasha [3.4K]
Im pretty sure the answer is distance and mass
4 0
2 years ago
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