We are given with
Cobalt phosphate - CoPO4
We are asked for the net ionic equation for the phosphate dissolving in H3O+
The net ionic equation is
CoPO4 (s) + H3O+ (aq) -----> HPO42- (aq) + Co3+ (aq) + H2O *(l)
Answer:
0.025 moles of NO₂ will produced
Explanation:
Given data:
Moles of NO₂ formed = ?
Volume of HNO₃ = 25.0 mL
Molarity of HNO₃ = 2 M
Solution:
Chemical equation:
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Number of moles of HNO₃:
Molarity = number of moles / volume in L
2M = number of moles / 0.025 L
Number of moles = 2 M × 0.025 L
Number of moles = 0.05 mol
Now we will compare the moles of HNO₃ with NO₂ from balance chemical equation.
HNO₃ : NO₂
4 : 2
0.05 : 2/4×0.05 =0.025
0.025 moles of NO₂ will produced.
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
The neutralization reaction is;
HClO₄(aq) + NaOH(aq) --> H₂O(l) + NaClO₄(aq)
Stoichiometry of HClO₄ to NaOH is 1:1
Therefore at equivalence an equal amount of NaOH reacts with an equal amount of HClO₄
The number of moles of NaOH in 1 L - 0.2 mol
Therefore in 21.52 ml of NaOH - 0.2/1000 x 21.52 = 0.004304 mol
the amount of HClO₄ moles that have reacted - 0.004304 mol
In 25 ml of HClO₄ - 0.004304 mol
Therefore in 1000 ml of HClO₄ - 0.004304/25 x 1000 = 0.172 mol
concentration of HClO₄ - 0.172 mol/L
