Answer:
a)
⇒
⇒
b)
⇒
⇒
Explanation:
A)
Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:
⇒
The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:
⇒
b)
Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:
⇒
Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:
⇒
Missing question:
Chemical reaction: H₂ <span>+ 2ICl → 2HCl + I</span>₂.
t₁ = 5 s.
t₂ = 15 s.
c₁ = 1,11 M = 1,11 mol/L.
c₂ = 1,83 mol/L.
rate of formation = Δc ÷ Δt.
rate of formation = (c₂ - c₁) ÷ (t₂ - t₁).
rate of formation = (1,83 mol/L - 1,11 mol/L) ÷ (15 s - 5 s).
rate of formation = 0,72 mol/L ÷ 10 s.
rate of formation = 0,072 mol/L·s.
Answer:
The differemt isotopes that differ in atomic mass
Explanation:
Answer:
4 × 10 g
Explanation:
Step 1: Write the balanced equation
2 H₂(g) + O₂(g) ⇒ 2 H₂O(I)
Step 2: Calculate the moles corresponding to 4 g of H₂
The molar mass of H₂ is 2.02 g/mol.
4 g × 1 mol/2.02 g = 2 mol
Step 3: Calculate the moles of H₂O produced from 2 moles of H₂
The molar ratio of H₂ to H₂O is 2:2. The moles of H₂O produced are 2/2 × 2 mol = 2 mol.
Step 4: Calculate the mass corresponding to 2 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
2 mol × 18.02 g/mol = 4 × 10 g
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over