Answer:
The answer to your question is P2 = 170.9 torr
Explanation:
Data
Volume 1 = 12.1 l Volume 2 = 21.1 l
Temperature 1 = 241 °K Temperature 2 = 298°K
Pressure 1 = 546 torr Pressure 2 = ?
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = T1V1T2 / T1V2
-Substitution
P2 = (241 x 12.1 x 298) / (241 x 21.1)
-Simplification
P2 = 868997.8 / 5085.1
-Result
P2 = 170.9 torr
Answer I'm not quite sure if this is right, but maybe it's because Jessica put her thermometer deeper in the stream, and Javier put his more up.
Explanation:
Under standard temperature and pressure conditions, it is known that 1 mole of a gas occupies 22.4 liters.
From the periodic table:
molar mass of oxygen = 16 gm
molar mass of hydrogen = 1 gm
Thus, the molar mass of water vapor = 2(1) + 16 = 18 gm
18 gm of water occupies 22.4 liters, therefore:
volume occupied by 32.7 gm = (32.7 x 22.4) / 18 = 40.6933 liters
Answer:
2.18×10⁴ kJ are liberated when you burn 0.554 kg of olive oil
Explanation:
The chart indicates that ΔH combustion for the olive oil is 39.31 kJ/g which means that, when you burn 1 g of olive oil, you have 39.31 kJ of heat.
In this case, the mass of olive is 0.554 kg. Let's convert the mass from kg to g → 0.554 kg . 1000 g / 1kg = 554 g
Now, the conversion factor:
554 g . 39.31 kJ / 1g = 2.18×10⁴ kJ
<span>IF WE TAKE N=1 IT IS CALLED GROUND STATE. THEN THE OTHER FOLLOWING HIGHER STATES ARE CALLED EXCITED STATES. IF THE ELECTRON IN AN ATOM JUMPS FROM A STATE TO A LOWER STATE, IT LOSES ENERGY. FROM THE GIVEN STATEMENT, THE WAY TO FIND THE ENERGY RELEASED IS GIVEN BY THE FORMULA, E(n)=(-13.6 eV)/n^2. FIRST TO FIND E(5)=(-13.6 eV)/(5)^2, WE GET E(5)=-0.544 eV. E(3)=(-13.6 eV)/(3)^2, WE GET E(3)=-1.5111 eV. THEN WE HAVE TO FIND THE ENERGY TRANSITION LEVEL. ON SUBTRACTING WE GET 0.967eV. THIS ENERGY HAS TO BE CONVERTED IN JOULES. SO WE MAKE E=0.967*(1.60*10^(-19)) J/eV, WHICH CORRESPONDS TO 0.15472*10^(-18) J. WE NEED TO FIND TO THE WAVELENGTH. THE CORRESPONDING FORMULA E = hf = hc/λ, λ = hc/E. BY SUBSTITUTING THE KNOWN VALUES, WE GET THE ANSWER TO BE 1285.548 NM.</span>
