All categories

3 years ago

QUESTION 3

Chemistry

1 answer:

nikitadnepr [17]3 years ago

7 0

Explanation:

Other important behaviors of gases explained by the Kinetic Molecular Theory are effusion and diffusion. Effusion is the rate at which a gas escapes through a small hole in a container. Diffusion is the rate at which a gas travels across a room. Both of these phenomena are illustrated by the following figure.

You might be interested in

Please help due in 5 min.

Lina20 [59]

Answer:

B) Hope this helps! :)

Explanation:

6 0

3 years ago

Consider the following reaction. SO2Cl2 → SO2 + Cl2. After collecting experimental data you found that plotting ln[SO2Cl2] vs. t

Nikitich [7]

Answer:

$[SO_2Cl_2]_{600}= 0.0842 M$

Explanation:

Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to $SO_2Cl_2$ .

The linear equation has the following terms:

$y = -0.000290t - 2.30$

It is a linear form of the integrated first-order law equation:

$ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o$

Therefore, the rate constant, k, is:

$k = 0.000290 s^{-1}$

The natural logarithm of initial molarity is:

$ln[SO_2Cl_2]_o = -2.30$

Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:

$ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474$

Take the antilog of both sides to find the actual molarity:

$[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M$

4 0

3 years ago

Methyl salicylate, C8H8O3, the odorous constituent of oil of wintergreen, has a vapour pressure of 1.00 torr at 54.3oC and 10.0

abruzzese [7]

<span>Use the Arrhenius equation. Use p1 and p2 and T1 and T2 and solve for Ea (actgivation energy) in Joules, then plug that back into the Arrhenius equation and either p1 or p2 to calculate p at 25C.</span>

5 0

3 years ago

1(30%). A thin plate black body, insulated at the bottom is placed faced up in a room with wall temperatures of 30 C. Air at 0 C

ch4aika [34]

Explanation:

The given data is as follows.

Heat transfer coefficient (h) = 12 $W/m^{2}.C$

Plate temperature ( $T_{1}$ ) = $30 ^{o}C$ = 303 K

Steady state temperature ( $T_{2}$ ) = ?

Hence, formula applied for steady state is as follows.

$(h \times A \times \Delta T)_{air}$ = $\sigma \times A \times (T^{4}_{1} - T^{4}_{2})_{plate}$

Putting the given values into the above formula as follows.

$(h \times A \times \Delta T)_{air}$ = $\sigma \times A \times (T^{4}_{1} - T^{4}_{2})_{plate}$

$12 \times (T_{2} - 273)$ = $5.67 \times 10^{-8} \times [(30 + 273)^{4} - T^{4}_{2}]$

$T_{2}$ = 282.66 K

= (282.66 -273) $^{o}C$

= 9.66 $^{o}C$

Thus, we can conclude that the steady state temperature will be 9.66 $^{o}C$ .

3 0

3 years ago

Which is easier to remove: a valence electron from, Li or Na? Explain why.

scZoUnD [109]

The atomic number of Li is 3

Electron configuration of Li : 1s² 2s¹

The atomic number of Na is 11

Electron configuration of Na : 1s²2s²2p⁶3s¹

Thus there is one electron in the valence shell of Li (2s¹) and that of Na (3s¹). However, the valence electron in Na is in a shell that is farther away from the nucleus compared to that of Li. As a result, the Na valence electron will be held less tightly by the nucleus i.e. it will experience a reduced nuclear attraction and can be removed easily than the Li 2s electron.

7 0

3 years ago