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Ksenya-84 [330]
3 years ago
7

Which of the following is/are part of the nitrogen cycle?

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
7 0
Denitrification is part of the nitrogen cycle 
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The one that will dissolve most quickly is : D. a tablespoon of ultrafine sugar

Unlike any other type of sugar, ultrafine sugar tend to have smaller crystal size. Which means that this substance will dissolve easier if it put into liquid such as tea or water

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According to Dalton, atoms cannot be SELECT ALL THAT APPLY a Created b Destroyed c Subdivided d Bonded​
Sveta_85 [38]

Answer:

The answer is subdivided,created or destroyed

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2 years ago
What would be a likely climatic outcome of rising ocean temperatures?
Kay [80]
I think your answer is D. However I’m not 100% sure.
7 0
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]

<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
3 years ago
You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th
ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially,  only 4 moles of oxygen gas were present in the flask.

p_{O_2}=Tp_1\times X_{O_2}  (X_{O_2}=\frac{4}{4}) ( according to Dalton's law of partial pressure)

p_{O_2}=Tp_1\times 1=Tp_1....(1)

Tp_1= Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

Tp_2= Total pressure of the mixture.

from (1)

Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}

On rearranging, we get:

Tp_2=2\times Tp_1

The new total pressure will be twice of initial total pressure.

7 0
3 years ago
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