when sodium metal is dropped in water, hydrogen gas in liberated due to extreme heat released as the reaction is exothermic, gas catches fire.
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Answer:
8.7 L
Explanation:
T2(V1/T1) = V2
417.15 K(6.2 L/296.45 K) = 8.7 L
Remember to almost always change celcius to kelvin. Also, this is part of Charle's Law (temp and volume are proportional, so if temp increaces so must the volume or vice versa). Lastly, Charle's Law has the formula of V1/T1 = V2/T2. I just rearranged it to go along with your problem. Hence, the T2(V1/T1) = V2
Answer:
Mass = 1274 .64 g it would be option C if it is converted into kilogram
1274 .64 / 1000 = 1.27 Kg
Explanation:
Given data:
Number of moles of C₂₀H₄₂ = 4.52 mol
Molar mass of carbon = 12 g/mol
Molar mass of hydrogen = 1.0 g/mol
Mass of C₂₀H₄₂ = ?
Solution:
Number of moles = mass / molar mass
Molar mass = 20× 12 + 42× 1.0 = 282 g/mol
Now we will put the values in formula:
Number of moles = mass / molar mass
4.52 mol = mass / 282 g /mol
Mass = 4.52 mol × 282 g/mol
Mass = 1274 .64 g
Answer:
Explanation:
Calcium is all around us. The average human contains approximately 1kg of calcium, of which 99% is stored in our bones. It is the 5th most abundant element in the earth's crust, occurring widely as calcium carbonate which is more commonly known as limestone. It is also the fifth most abundant dissolved ion in seawater.
