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2 years ago

Explain why in the early morning just before dawn breaks, the temperature of the air

Chemistry

1 answer:

Oksanka [162]2 years ago

7 0

Answer:

Once the sun comes up, the sunlight excites the cold air in the first foot or so above the ground (which can be 10 or more degrees colder), which causes it to move around and mix into the next several feet of air. That “mixing upward” drops the temperature of the air at thermometer level.

Explanation:

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Which of these is most likely to create large areas of land subsidence?

Mumz [18]

I think it is Global warming

6 0

2 years ago

If acetic acid is the only acid that vinegar contains (ka=1.8×10−5), calculate the concentration of acetic acid in the vinegar.

kicyunya [14]

Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

$CH_{3}COOH \ \textless \ ---\ \textgreater \ H^{+} + CH_{3}COO^{-}$

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)

pH = 2.88 ==> $[H^{+}]$ = $10^{-2.88}$ = 0.001 $moldm^{-3}$

The change in Concentration Δ $[CH_{3}COOH]$ = 0.001 $moldm^{-3}$

  CH3COOH H+ CH3COOH
Initial   $moldm^{-3}$      $x$ 0 0

Change $moldm^{-3}$ -0.001 +0.001 +0.001

Equilibrium $moldm^{-3}$ $x$ - 0.001 0.001 0.001


Since the $k_{a}$ value is so small, the assumption
$[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}$ can be made.

$k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}$

Solve for x to get the required concentration.

note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.

2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind

Hope this helps!

8 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,

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

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

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

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

$E_a=33813.28\ J/mol$

The activation energy for the decomposition = 33813.28 J/mol

8 0

2 years ago

What process is being shown by water being given off from each bond site?

34kurt

The process that is being shown by water being given off from a bond site is DEHYDRATION SYNTHESIS.
Dehydration synthesis is the process of joining two molecules or compounds together as a result of removal of water.

4 0

3 years ago

A car travels at a speed of 80 km/h. Which of the following is correct to convert the speed of the car to hm/s? [1 km = 10 hm, 1

amm1812

Answer: The correct answer is Option D.

Explanation:

We are given that car travels with a speed of 80 km/hr and we need to convert it into hm/s.

For that we use the conversions:

1 kilometer = 10 hectometer

And, 1 hour = 3600 seconds

To convert the quantity into hm/s, we multiply the numerator which is 80 by 10 to convert the whole quantity into hm/hr and now, to convert the result into hm/s finally, we divide the result by 3600.

$\Rightarrow (\frac{80km1}{hr})(\frac{10hm}{1km})=\frac{80km}{hr}\\\\\Rightarrow (\frac{80km}{hr})(\frac{1hr}{3600s})=\frac{80hm}{3600s}$

Hence, the correct answer is option D.

4 0

2 years ago