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Nezavi [6.7K]
3 years ago
6

Due Today!!!

Chemistry
1 answer:
Taya2010 [7]3 years ago
5 0
Answer B hope this helps
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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]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

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

8 0
2 years ago
Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?
rodikova [14]
<span>Equation:2H2(g) + O2(g) → 2H2O(g)
</span><span>
Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

</span>
4 0
2 years ago
Read 2 more answers
If the mole fraction of nitric acid (HNO3) in an aqueous solution is 0.275, what is the percent by mass of HNO3?
jeka94
Molar mass of    N    = 14 g/molMolar mass of   O2    = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2)                                                                                             = N3O6So according to above explanation,D) N3O6, is the correct answer.

8 0
2 years ago
Metals tend to ________ electrons and form _______ ions.
masya89 [10]
<span>Metals tend to lose electrons and form electro-positive ions / cations.</span>
5 0
3 years ago
The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g
ANTONII [103]

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

7 0
3 years ago
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