This compound is also known as Calcium Oxide. The SI base unit for amount of substance is the mole. 1 grams CaO is equal to 0.017832495800447 mole. Note that rounding errors may occur, so always check the results.
Hope I helped :)
Answer: CrO₄⁻ and Ba²⁺
Explanation:
1) Chemical equation given:
2H⁺ + CrO₄⁻ + Ba²⁺ + 2OH⁻ → Ba²⁺ + CrO₄⁻ + 2H₂O
2) Analysis
That is an oxidation-reduction equation (some species are been oxidized and others are being reduced).
The given equation is known as total ionic equation, because it shows all the species as ions that are part of the reaction.
2) Specator ions
Spectator ions are the ions that do not change their oxidation state and are easily identified as they are the same in the reactant and product sides.
Here the ions that are the same in the reactant and product sides are:
CrO₄⁻ and Ba²⁺
3) Addtitional explanation.
Once you identify the spectator ions you can delete them from the equation to obtain the net ionic equation , which in this case turns to be:
2H⁺ + 2OH⁻ → 2H₂O
But this is not part of the question; it is some context to help you understand the use of the spectator ions concept.
We are asked to convert 25 cg to units of hg.
1 cg = 1 centigram = 10⁻² g
1 hg = 1 hectogram = 10² g
The options given are:
a) 1 hg/ 10² g
b) 10² cg/ 1 hg
c) 10² hg/ 1 cg
d) 10⁻² g/ 1 cg
To convert 25 cg to 1 hg, we could convert the 25 cg to grams first, then grams to hg.
25 cg · 10⁻² g/ 1cg = 0.25 g
Here we have converted our number from cg to grams. We can use another conversion of grams to hg to complete the conversion.
0.25 g · 1 hg/ 10² g = 0.0025 hg
Therefore, the first conversion we used was d) 10⁻² g/ 1 cg.
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
I think ita c sense its the same object