Answer:
Explanation:
Oxidation no is equal to charge on each atomic ion. If it is increased , element is oxidised and if it is decreased , element is reduced .
2AgCl+Zn⟶2Ag+ZnCl2
Zinc is oxidised , Ag is reduced .
Ag⁺ converts to Ag . ( oxidation number is reduced ) so Ag is reduced.
Zn converts to Zn⁺² ( oxidation number is increased ) so Zn is oxidised .
4NH₃+3O₂⟶2N₂+6H₂O
oxidation number of nitrogen in ammonia is - 3
oxidation no of nitrogen in nitrogen is zero.
Oxidation no of nitrogen is increased so it is oxidised.
oxidation no of oxygen is zero in oxygen and its oxidation no in water is -2 . So oxidation no is reduced so oxidation is reduced.
Fe₂O₃+2Al⟶Al₂O₃+2Fe
oxidation no of Fe in Fe₂O₃ is + 3 and it is zero in Fe so iron is reduced.
oxidation no of Al in Al is zero and it is +3 in Al₂O₃ so it is oxidised .
It matches the universal pH indicator and is indicating the proper pH
Answer:
(a) I⁻ (charge 1-)
(b) Sr²⁺ (charge 2+)
(c) K⁺ (charge 1+)
(d) N³⁻ (charge 3-)
(e) S²⁻ (charge 2-)
(f) In³⁺ (charge 3+)
Explanation:
To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.
(a) |
I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).
(b) Sr
Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).
(c) K
K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).
(d) N
N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).
(e) S
S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).
(f) In
In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).
Answer:
The partial pressure of CO2 is 712,8 in torr
Explanation:
Molar fraction = Pressure in a compound / Total Pressure
Molar fraction H20 = 21,2 / 734 = 0,0288
Sum of molar fraction in a sample = 1
1 - 0,0288 = 0,9712 (molar fraction of CO2)
Molar fraction CO2 = Pressure CO2 / Total pressure
0,9712 . 734 = Pressure CO2
712,8 =Pressure CO2
Answer:
The value of the silver in the coin is 35.3 $
Explanation:
First of all, let's calculate the volume of the coin.
2π . r² . thickness = volume
r = diameter/2
r = 41 mm/2 = 20.5 mm
2 . π . (20.5 mm)² . 2.5 mm = 6601 mm³
Now, this is the volume of the coin, so we must find out how many grams are on it.
6601 mm³ / 1000 = 6.60 cm³
Let's apply density.
D = Mass / volume
10.5 g/cm³ = mass /6.60 cm³
10.5 g/cm³ . 6.60 cm³ = mass
69.3 g = mass
Each gram has a cost of 0.51$
69.3 g . 0.51$ = 35.3 $
