1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
Answer:
27.0
Explanation:
Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.
Answer:
Sodium arachidate; Sodium palmitate and Sodium palmitate
Explanation:
Triglycerides are esters of fatty acids with glycerol. In triglycerides, three fatty acid molecules are linked by ester bonds to each of the three carbon atoms in a glycerol molecule. The fatty acids may be same or different fatty acid molecules. Hydrolysis of triglycerides yields the three fatty acid molecules and glycerol.
Saponification is the process by which a base is used to catalyst the hydrolysis of the ester bonds in glycerides. The products of this base-catalyzed hydrolysis of triglycerides are the metallic salts of the three fatty acids and glycerol. The salts of the fatty acids are known as soaps.
For a triglyceride that has the fatty acid chains arachidic acid, palmitic acid and palmitic acid attached to the three backbone carbons glycerol, the saponification of the triglyceride with NaOH will yield the sodium salts or soaps of the three fatty acids as well as glycerol.
Arachidic acid will react with NaOH to yield sodium arachidate.
The two palmitic acid molecules will each react with NaOH to yield sodium palmitate.
It would protect best against C) Alpha radiation, as Beta radiation is stopped by lighter metals such as aluminium, and Gamma radiation can only be stopped by heavier metals such as lead.
Answer:
a. 82.68 g b. 9.8 min
Explanation:
a. The amount of gold deposited by 5 A current in 2 hrs 15 mins
Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s
Q = 5 A × 8100 s
= 40500 C
Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C
n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole
Now n = m/M where m = mass of gold and M = molar mass of gold = 197
m = nM
= 0.42 × 197 g
= 82.68 g
b. The time taken for 6g of gold to be deposited.
We first find the number of moles of gold in 6g of gold
Since n = m/M and m = 6 g
n = 6/197 = 0.0305 mole
Q = It = nF
t = nF/I
= 0.0305 mol × 96500 C/5 A
= 2939.09 mol C
= 587.82 s
Changing t to minutes
587.82/60 s = 9.8 min
