Answer:
10 atm
Explanation:
There's a lot to do here, but lets take it one step at a time. First, let's write a balanced equation for the decomposition of potassium chlorate into potassium chloride and oxgyen gas.
2 KClO3 → 2 KCl + 3 O2
Now let's find the moles of the KClO3 (molar mass 122.55 g/mol) that we have take 10 g/122.55 g/mol, grams will cancel and we are left with 0.0816 moles. lets divide that by two since we have a two in front of the KClO3 in the equation, and then multiply that number by 5 since it's the total moles of products, in summary, multiply by 5/2 to get 0.204 moles.
Now that we know the moles of our products, let's plug some stuff into the ideal gas law PV = nRT. We are looking for P so let's solve for that. P = (nRT)/V, now let's plug in our values. Make sure V is converted to liters so 0.5 L. And convert celcius to kelvin by adding 273
P = ((0.204 moles)(318 K)(0.08206 L atm mol^-1 K^-1))/0.5 L
A lot of units cancel, and we get about 10.65 atm, if you don't want the answer in atm, you can find a conversion equation. But let's round to sig figs for now, which will bring us to 10 atm.
Answer:
b. 485 kPa
Explanation:
Gay-Lussac's law express that the pressure of a gas under constant volume is directly proportional to the absolute temperature. The equation is:
P1T2 = P2T1
<em>P is pressure and T absolute temperature of 1, initial state and 2, final state of the gas</em>
<em>Where P1 = 74psi</em>
<em>T2 = 20°C + 273.15 = 293.15K</em>
<em>P2 = ?</em>
<em>T1 = (95°F -32) * 5/9 + 273.15 = 308.15K</em>
<em />
Replacing:
74psi*293.15K = P2*308.15K
70.4psi
In kPa:
70.4psi * (6.895kPa / 1psi) =
<h3>b. 485 kPa
</h3>
Answer:
.. / -.. --- -. - / -.- -. --- .-- / .-- .... .- - / .. / .-- .- ... / - .... .. -. -.- .. -. --. / .-.. . .- ...- .. -. --. / -- -.-- / -.-. .... --- .. .-.. -.. / -... . .... .. -. -.. / -. --- .-- / .. / ... ..- ..-. ..-. . .-. / .- / -.-. ..- .-. ... . / .- -. -.. / .. / .- -- / -... .-.. .. -. -.. / - .... .-. ..- / .- .-.. .-.. / - .... .. ... / .- -. --. . .-. / --. ..- .-.. - / ... .- -.. -. . ... ... / -.-. --- -- .. -. --. / - --- / .... .- ..- -. - / -- . / ..-. --- .-. . ...- . .-. / .. / -.-. .- -. - / .-- .- .. - / ..-. --- .-. / - .... . / -.-. .-.. .. ..-. ..-. / .- - / - .... . / . -. -.. / --- ..-. / - .... . / .-. .. ...- . .-. .-.-.- / .. ... / - .... .. ... / .-. . ...- . -. --. . / - .... .- - / .. -- / ... . . -.- .. -. --. ..--.. / --- .-. / ... . . -.- .. -. --. / ... --- -- --- -. . / - --- / .- ...- . -. --. . / -- . ..--.. / ... - ..- -.-. -.- / .. -. / -- -.-- / --- .-- -. / .--. .- .-. .- -.. --- -..- / .. / .-- .- -. .- / ... . - / -- -.-- / ... . .-.. ..-. / ..-. .-. . . .-.-.-
Explanation:
Answer: 1. halve
2. halve
3. double
Explanation:
The relationship between wavelength and energy of the wave follows the equation:
E= energy
= wavelength of the wave
h = Planck's constant
c = speed of light
Thus as wavelength and energy have inverse realation, when wavelength will halve , energy will double.
2. The between wavenumber and energy of the wave follows the equation:
E= energy
= wavenumber of the wave
h = Planck's constant
c = speed of light
Thus as wavenumber and energy have direct relation, when wavenumber will halve , energy will be halved.
3. The relationship between energy and frequency of the wave follows the equation:
where
E = energy
h = Planck's constant
= frequency of the wave
Thus as frequency and energy have direct realation, when frequency will double , energy will double.
Explanation:
<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these cases</u></em>
<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these caseshope it helps you like me plz</u></em>
