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3 years ago

How many particles are in 8.00 moles of bromine?

Chemistry

2 answers:

tester [92]3 years ago

7 0

Answer:

639.232 grams

Explanation:

Jet001 [13]3 years ago

4 0

Answer: 4.816 x 10^24 Particles

Explanation: 1mole of any substance contains 6.02x10^23 Particles

Therefore, 8 moles of bromine will contain = 8 x 6.02x10^23 = 4.816 x 10^24 Particles

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How many molecules of NaOH are in 23 moles of NaOH

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120 grams of NaOH means 3 Moles of NaOH because 40 grams (Molecular Weight in grams) is one mole of NaOH.

Each mole of any substance contain Avogadro’s number of molecules ie., 6.022 x (10 to the power 23).

Hence 3 Moles of NaOH contain 3 times of Avogadro’s number of molecules ie., 3 x 6.022 x (10 to the power 23)

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3 years ago

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3 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

Answer: The freezing point of solution is 2.6°C

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

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3 years ago