Balanced equation: 2Fe + 3H2O → Fe2O3 +3H2
Convert g to mols:
285/55.845 = 5.1034 mols
Mole ratio of Iron and Iron (III) Oxide: 2:1
5.1034/2 = 2.5517 mols
C6H12O + 6OC2 + 6H2O + energy
Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L
As per the evenly distributed response
2NaCl (g) ----> 2Na(l)+ Cl2(g)
Calculate the amount of Cl2 that was formed as indicated below:
Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)
= 673.9 mol
P is equal to 1.0 atm, and T is equal to 813.15 K
when converted to Kelvin by multiplying by a factor of 273.15.
Using Cl2 as an ideal gas, determine the in the following volume:
volume = nRT/P
= 673.9 * 0.0821 * 813.15/ 1
=4.5×10^4 L
As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L
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Answer:
Explanation:
Salt water intrusion can cause the <u><em>fresh</em></u> water in wells to become contaminated with<u><em> salt</em></u>water.
Answer:
A. Cu^+2(aq)cathode ---> Cu^+2(aq)anode
Explanation:
Electrolysis is the process in which current is passed through a solution thereby causing a chemical change at the anode and cathode. Copper is being purified using electrolysis by using impure copper at the anode and pure copper at the cathode. This pure and impure copper are placed in a copper(ii)sulfate electrolyte solution and dc current is made to pass through it. The resulting changes at the anode and cathode are given by the equation:
cathode: Cu²⁺ + 2e⁻ ⇒ Cu
anode: Cu ⇒ Cu²⁺ + 2e⁻
