For NaBr(aq) hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. For NaBr(aq), bromine, Br2(l), will be produced instead of oxygen gas at the anode.
<span>For sodium sulfate hydrogen gas is formed at the cathode. Hydrogen in water is reduced to hydrogen gas. Oxygen gas will be produced at the anode. </span>
<span>Should someone suggest that sodium metal is formed at the cathode, rest assured that that can't happen. Sodium metal reacts with water to make Na+. </span>
<span>Lead(II) iodide is insoluble in water. Therefore, not much will happen since there is no electrolyte.</span>
Answer:
Explanation: The lowest pressure in a laboratory is 4.0×10^-11Pa
Using Ideal gas equation
PV = nRT
P= 4.0×10^-11Pa
V= 0.020m^3
T= 20+273= 293k
n=number of moles = m/A
Where m is the number of molecules and A is the Avogradro's number=6.02×10²³/mol
R=8.314J/(mol × K)
PV= m/A(RT)
4.0×10^-11 ×0.020 = m/6.02×10²³(8.314×293)
m = 4.0×10^-11×0.020×6.02×10^23 / (8.314×293)
m = 1.98×10^8 molecules
Therefore,the number of molecules is 1.98×10^8
Answer:
An EL mechanism
Explanation:
I looked this up, it's called an EL mechanism
Answer:
123.2 Liters.
Explanation:
At STP (T = 273K & P = 1atm)<em>, one mol of any gas will occupy 22.4 liters</em>.
With the above information in mind, we can <u>calculate how many liters would 5.500 mol of gas occupy</u>:
5.500 mol * 22.4 L / mol = 123.2 L
So 5.500 moles of C₃H₃ would have a volume of 123.2 liters at STP.
B i think but im not 100 percent sure
