Answer:
Sulfur would gain electrons
Explanation:
Atoms want to have a complete out valence shell and because sulfur only needs 2 more electrons to complete the outer shell it would take 2 more.
Answer:
One of the main uses of the cathode ray tube is in the Cathode ray oscilloscope
Explanation:
Cathode rays are produced when a gas in an evacuated glass at very low pressure and high pressure breaks up into positive and negative ions. the negative ions move towards the anode(positive electrode) while the positive ions move towards the cathode(negative electrode), and there they knock off electrons (which are known as cathode rays) from the metal plate of the cathode.
Cathode ray tubes are mainly used in oscilloscopes, television picture tubes and in computer screens.
The cathode ray oscilloscope is used in a.c. and d.c. voltage measurements, observation of waveforms, time measurements, etc.
Answer: S8(s) + 8O2(g) → 8SO2(g) ΔH = -2368 kJ a.
The accepted model of the atom was changed.
Answer:
ΔG° of reaction = -47.3 x
J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x
M
[ATP] = 1.2 x
M
[ADP] = 8.4 x
M
Let's plug in these values in the above equation for equilibrium constant:
K = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
K = 1.47 x
M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5
) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x
) + (-16810.68)
ΔG° of reaction = -47.3 x
J/mol