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3 years ago

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What is the density of krypton gas (MM = 83.8 g/mol) at 3.00 atm and 100. °C? g/L

1 answer:

stiks02 [169]3 years ago

3 0

Answer:

8.21

Explanation:

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I have a brick, dimensions 4 cm tall, 9 cm wide, and 25 cm long, with a mass of 1250 g. What is the density?

Likurg_2 [28]

Answer:

1.389

Explanation:

P = m / V

m = 1250

V = 4x9x25 = 900

1250/900 = 1.388889

7 0

3 years ago

How are scientific questions answered?

galben [10]

Answer:

D

Explanation:

Scientists always perform tests and observe and measure in the physical world to prove their points or answer their questions.

3 0

3 years ago

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =

Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = ?

$k_2=4.0\times 10^3s^{-1}$

$k_1=1.5\times 10^3s^{-1}$

$T_1=5\ ^0C$

$T_2=25\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (5 + 273.15) K = 278.15 K

T = (25 + 273.15) K = 298.15 K

$T_1=278.15\ K$

$T_2=298.15\ K$

So,

$\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)$

$E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}$

$E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}$

$E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}$

$E_a=33813.28\ J/mol$

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0

3 years ago

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

<u>As we know</u>–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

2 years ago

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Triss [41]

Answer:

o

Explanation:

6 0

2 years ago