Answer:
Explanation:
1) Vanadium (IV) → V⁺⁴
Carbonate → CO₃⁻²
So , Vanadium (IV) Carbonate = V₂(CO₃)₄ or V(CO₃)₂
2) Tin (II) = Sn⁺²
Nitrite = NO₂⁻
So, Tin (II) Nitrate = Sn(NO₂)₂
3) Cobalt (III) = Co⁺³
Oxide = O⁻²
So , Cobalt (III) Oxide = Co₂O₃
4) Titanium (II) = Tn⁺²
Acetate = CH₃COO⁻
So , Titanium (II) Acetate = Tn(CH₃COO)₂ or Tn(C₂H₃O₂)₂
5) Vanadium (V) = V⁺⁵
Sulfide = S⁻²
So , Vanadium (V) Sulfide = V₂S₅
6) Chromium (III) = Cr⁺³
Hydroxide = OH⁻
So , Chromium (III) Hydroxide = Cr(OH)₃
7) Lithium = Li⁺
Iodide = I⁻
So , Lithium Iodide = LiI
8) Lead (II) = Pb⁺²
Nitride = N⁻³
So , Lead (II) Nitride = Pb₃N₂
9) Silver = Ag⁺
Bromide = Br⁻
So , Silver Bromide = AgBr
Molarity of solution is defined as number of moles present in one liter solution. Mathematically, it is expressed as
Molarity =

Thus, if 1 mole of solute is present in 1 liter solution, molarity of solution is 1 M.
In present case, initial conc, of solution was 1.25 M.
∴ Number of moles of <span>co[h2o]6cl2 available initially = 1.25 mole, if the solution is 1 liter</span>
A) 4400 kj of heat released into surroundings
<h3>Further explanation</h3>
Reaction
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O, the ∆H is –2200 kJ
Reaction exothermic( ∆H=-, released heat to surrounding) and for combustion of 1 mole of C3H8
So for two moles of C3H8, the enthalpy :

Answer:
A (contains most of the mass of the atom)
Evidence has it that a proton is about 2000 times as massive as an electron.
And there is usually multiple protons and neutrons in the nucleus
From what I just said, you can say that B is wrong
C however is also wrong because protons have a +charge and neutrons are neutrle which means you always have a charge > (greater than) 0
And D is wrong because electrons (which are not in the nucleus) have a neg charge. and protons have a + charge and are in the nucleus
So your answer is A
Hope it helped
Spiky Bob