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2 years ago

The nucleus of the cell

Chemistry

1 answer:

Stells [14]2 years ago

3 0

Answer:

Explanation:

it is c because that is the andwer

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Write a balanced equation for the following:

Ket [755]

(ANS1)— P4 + 5O2 ---> 2P2O5

(ANS2)— C3H8 + 5O2---> 3CO2 + 4H20

(ANS3)— Ca2Si + 4Cl2 ---> 2CaCl2 + SiCl4

7 0

2 years ago

The temperature of nitrogen is unknown when the gas occupies 100.0 mL at 99.10 kPa. If the gas is known to occupy 74.2 mL at 133

laiz [17]

Answer:

The unknown temperature is 304.7K

Explanation:

V1 = 100mL = 100*10^-3L

P1 = 99.10kPa = 99.10*10³Pa

V2 = 74.2mL = 74.2*10^-3L

P2 = 133.7kPa = 133.7*10³Pa

T2 = 305K

T1 = ?

From combined gas equation,

(P1 * V1) / T1 = (P2 * V2) / T2

Solving for T1,

T1 = (P1 * V1 * T2) / (P2 * V2)

T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)

T1 = 3022550 / 9920.54

T1 = 304.67K

T1 = 304.7K

3 0

2 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$