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Marina86 [1]
3 years ago
10

In addition, they are poor conductors of

Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer: electricity

Explanation:

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Urea makes up a significant amount of urine, and is the chemical that the body uses to rid cells of unwanted carbon dioxide and
Ira Lisetskai [31]
Urea is highly soluble in water. When it is allowed to dissolve in water in the presence of heat, it will yield ammonia and carbon dioxide. The reaction is shown below:

<span>NH2-CO-NH2 + H2O </span>⇒ 2 NH3 + CO2

As you can observe in the stoichiometric equations, 1 molecule of water can dissolve with 1 mole of urea.
3 0
3 years ago
It is necessary to have a 40% antifreeze solution in the radiator of a certain car. the radiator now has 70 liters of 20% soluti
Akimi4 [234]
The answer is : 17.5 liters drained and replaced by 17.5 liters of 100% solution. 
x = amount drained and replaced
(70-x) = remaining amount of 20% solution
<span>.20(70-x) + 1.00(x) = .40(70)
14 - .2x + 1x = 28
1x - .2x = 28 - 14
</span><span>.8x = 14
</span><span>x = 14/.8
x= 17,5 ( 17.5 liters drained and replaced by 17.5 liters of 100% solution)

</span>
6 0
2 years ago
Read 2 more answers
Which family contains elements with a full octet of valence electrons?
iren [92.7K]
D. the noble gasses !!!!
5 0
3 years ago
Read 2 more answers
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
Un buzo que nada en las profundidades despide pequeñas burbujas de aire¿Aumenta disminuye o permanece igual el tamaño de las bur
alekssr [168]

Answer:

Al ascender las burbujas van aumentando de tamaño.

Explanation:

Las burbujas que produce el buzo debajo del agua son pequeñas moléculas de dióxido de carbono gaseoso producto de la respiración del mismo.

Ahora, a medida que las burbujas suben a la superficie, la presión que sufren estas (Presión debido al agua), es menor conforme van ascendiendo debido a la ley de Boyle: A medida que la presión aumenta, el volumen va disminuyendo.

Esto significa que al ascender las burbujas van aumentando de tamaño debido a que la presión que sufren estas es menor que cuando están a mayores profundidades.

5 0
3 years ago
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