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WINSTONCH [101]
3 years ago
9

Does anyone know this???

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
4 0
It’s A! Have a great day!
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Answer:

D

Explanation:

John is not a very good businessman.

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What is the least polar solvent /solvent system used to run the tlc?
Lynna [10]
I believe that it is petroleum ether.
8 0
3 years ago
Consider the following equilibrium:
Ainat [17]

<u>Answer: </u>The equation which is wrong is K_p=K_c(RT)^{-5}

<u>Explanation:</u>

For the given reaction:

4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)

The expression for K_c\text{ and }K_p is given by:

K_c=\frac{1}{[O_2]^3}

K_p=\frac{1}{[O_2]^3}

The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.

Relationship between K_p\text{ and }K_c is given by the expression:

K_p=K_c\times (RT)^{\Delta n_g}

where,

\Delta n_g= number of moles of gaseous products - number of moles of gaseous reactants

R = gas constant

T= temperature

For the above reaction,

\Delta n_g = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3

Hence, the expression for K_p is:

K_p=K_c\times (RT)^{-3}

Therefore, the equation which is wrong is K_p=K_c(RT)^{-5}

7 0
3 years ago
Indicar la cantidad de sustancia en: a) 2.0L de un gas en C.N; b) 22.4 mL de un gas a 755 torr y 26 grados celsius?
nekit [7.7K]

Answer:

El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.

8 0
2 years ago
Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U
Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of nitrogen gas = 79mL/s

R_2 = rate of effusion of sulfur dioxide gas = ?

M_1 = molar mass of nitrogen gas  = 28 g/mole

M_2 = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}

R_2=52mL/s

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0
3 years ago
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