Answer: In CaF2, the oxidation number of Ca is +2
, and that of F is -1
. In H2SO4, the oxidation number of H is +1
, that of S is +6
, and that of O is -2
. In CaSO4, the oxidation number of Ca is +2
, that of S is +6
, and that of O is -2
. In HF, the oxidation number of H is +1
, and that of F is -1
Answer:
3Na2SO4
Explanation:
3Na2SO4
# of molecules: 3 moles of Na2SO4 or 3 × 6.22 × 10^23 molecules.
# of elements: 3 elements l
Name of element: = Sodium, S = Sulphur, O = Oxygen
# of atoms: Na = 6 atoms, S= 3, O= 12
Total # of atoms: 21
The #3 is a Coefficient.
Please mark branliest if you are satisfied. Thanking in anticipation.
Answer:
80cm3 of water, and 60cm3 carbon IV oxide is formed while 20cm3 of oxygen is left unreacted.
Explanation:
From Gay-Lussac's law, there are five volumes of oxygen, 1 volume if propane, 4 volumes of water and three volumes of CO2. Applying this shows the reacting volumes as we have in the image attached, hence the volumes left after reaction.
2(NH4)3PO4 (aq) + 3Ni(NO3)2(aq) ------> Ni3(PO4)2(s) + 6NH4NO3 (aq)
Ni3(PO4)2 is a precipitate.
Answer:
1. All red calves i.e. RR
2. All roan calves i.e RW
3. 2 red calves (RR) and two roan calves (RW)
Explanation:
According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):
1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR
2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW
3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).