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Ber [7]
2 years ago
6

Write insoluble or soluble solution for the following mixtores12345​

Chemistry
1 answer:
dimaraw [331]2 years ago
6 0

<em>Gasoline</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>

<em>acetone</em><em> </em><em>in</em><em> </em><em>nail</em><em> </em><em>polished</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>

<em>salt</em><em> </em><em>in</em><em> </em><em>alcohol</em><em> </em><em>=</em><em> </em><em>soluble</em><em> </em>

<em>oil</em><em> </em><em>in</em><em> </em><em>vinegar</em><em> </em><em>=</em><em> </em><em>insoluble</em><em> </em>

<em>tawas</em><em> </em><em>in</em><em> </em><em>water</em><em> </em><em>=</em><em> </em><em>soluble</em><em>.</em><em>.</em><em>.</em>

<em>Sorry</em><em> </em><em>if</em><em> </em><em>i</em><em> </em><em>am</em><em> </em><em>incorrect</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.
ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_A+p_B+p_C...

Given : p_{total} =total pressure of gases = ?

p_{O_2} = partial pressure of oxygen = 100 mm Hg

p_{N_2} = partial pressure of nitrogen = 573 mm Hg

p_{CO_2} = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

p_{H_2O} = partial pressure of water vapor = 47 torr = 47 mm Hg  (1torr=1 mm Hg)

putting in the values we get:

p_{total}=(100+573+40.28+47)mmHg

p_{total}=760.28mmHg

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

5 0
3 years ago
For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

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3 years ago
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Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
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No!! IT'S TRUE!!!!!

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