Answer:
You subtract the atomic number from the mass number to find the number of neutrons. If the atom is neutral, the number of electrons will be equal to the number of protons.
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
Use your periodic table to get 1 mol. Look up the atomic masses. I'll round the numbers since every periodic table is different.
N = 14
O2 = 2 *16
Total = 30 grams / mol
Therefore 2 mols = twice as much = 60 grams.
Answer:
c) HCl and NaCl
Explanation:
Since all the solutions are on a 1:1 mole ratio the comparison is straight forward.
The lowest pH will be for solution c) which has a strong acid, HCl, which ionizes 100 % and the neutral salt NaCl (which is neutral since it is derived from the reaction of the strong acid HCl and the strong base NaOH).
Solutions a) and b) are buffers of the weak base NH₃ and its conjugate acid NH₄⁺ and weak acid H₃PO₄ and its conjugate weak base NaH₂PO₄ respectively.
Solution c) is a basic solution being a mixture of the weak base NH₃ and the strong base NaOH
Solution e) is a mixture of a weak base NH₂ and weak acid HC₂H₃O₂
Answer:there is visible blue black coloration
Explanation:starch is a polysaccharide,it is composed of only glucose combined by glycosidic bonds.starch is majorly an insoluble carbonhydrate.
Starch consist of two distinguishable polysaccharide fraction namely amylose and amylopectin.
The iodine test is used to check for the presence of starch.
It gives a blue black coloration for amylose which is present in potato.
But in amylopectin,it gives a reddish coloration.
