Here, we use the mole as we would use any other collective number: a dozen eggs; a Bakers' dozen; a Botany Bay dozen.
Of course, the mole specifies a much larger quantity, and if I have a mole of stuff then I have
6.022
×
10
23
individual items of that stuff. We can also specify an equivalent mass, because we also know the mass of a mole of iron, and a mole of oxygen etc........The mole is thus the link between the macro world of grams and kilograms and litres, that which we can measure out in the lab, to the micro world of atoms, and molecules, that which we can perceive only indirectly.
Here we have the formula unit
F
e
2
(
S
O
4
)
3
. If there is a mole of formula units, there are necessarily 2 moles of iron atoms, 3 sulfate ions,.......etc.
Answer is: mass of the ore is 8.54kg.<span>
</span>ω(Ca₃(PO₄)₂ - calcium phosphate) = 58.6% ÷ 100% = 0.586.
m(P) = 1.00 kg · 1000 g/kg.
m(P) = 1000 g.
In one molecule of calcium phosphate there are two phosphorus atoms:
M(Ca₃(PO₄)₂) = 310.18 g/mol.
M(P) = 30.97 g/mol.
For one kilogram of phosphorus, we need:
M(Ca₃(PO₄)₂) : 2M(P) = m(Ca₃(PO₄)₂) : m(P).
310.18 g/mol : 61.94 g/mol = m(Ca₃(PO₄)₂) : 1000 g.
m(Ca₃(PO₄)₂) = 5007.75 g ÷ 1000 g/kg = 5.007 kg.
Mass of ore find from proportion:
m(Ca₃(PO₄)₂) : m(ore) = 56% : 100%.
m(ore) = 100% · 5.007 kg ÷ 58.6%.
m(ore) = 8.54kg.
This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
Learn more:
Answer:
true
Explanation:
This is because it helps distribute the particles that are being dissolved.
Answer:
wouldn't it be the #2 because +∆H um I forgot
