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3 years ago

can someone please tell my why my answer is incorrect and what the correct answer is? I don’t understand how this is wrong

Chemistry

1 answer:

forsale [732]3 years ago

6 0

Answer:

you were correct from my point of view but you could add a few zero's

Explanation:

i don't see how the was wrong

33000 Grams/Milliliters (g/mL) = 33,000,000 Milligrams/Milliliter (mg/mL)

1 g/mL = 1,000 mg/mL

1 mg/mL = 0.001000 g/mL

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The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

3 0

3 years ago

Si una masa dada de hidrógeno ocupa 40 litros a 700 grados torr. ¿Qué volumen ocupará a 1 atmósfera de presión? (dar la presión

Svetach [21]

Answer:

$V_2=36.84L$

Explanation:

Hola,

En este caso, podemos usar la ley de Boyle, la cual nos permite analizar el comportamiento volumen-presión en un gas ideal de manera inversamente propocional:

$P_1V_1=P_2V_2$

Así, dado el volumen y la presión inicial, la cual se convierte a atmósferas (760 torr = 1atm), calculamos el volumen final a 1 atm como se muestra a continuación:

$V_2=\frac{P_1V_1}{P_2}=\frac{700torr*\frac{1atm}{760torr}*40l }{1atm}\\ \\V_2=36.84L$