Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
For an non spontaneous reaction between silver (Ag) and copper (Cu) and their ions, Cu is the oxidizing agent while Ag+ is the reducing agent,
The following reactions will take place;
Anode Cu = Cu+2 + 2e- E= +0.34 volts
Cathode; Ag+ + e = Ag E = +0.80 volts
The net reaction will be Cu + 2Ag+ = Cu+2 + 2Ag
Thus, the voltage will be
= +0.80 - (+0.34)
Answer:
2NaBr + I2 Right arrow. 2NaI + Br2
2AgNO3 + Ni Right arrow. Ni(NO3)2 + 2Ag
Explanation:
The activity or electrochemical series is an arrangement of elements according to their order
of reactivity.
If we look at the reactions, one thing that we must note is that the reactions that can occur are those in which an element that is higher in the series displaces another element that is lower in the series.
Br is higher in the electrochemical series than I so it can displace it. Ni is higher than Ag in the electrochemical series hence it can displace it.
590 mL = 590 cm³= 0,59 dm³
C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________
M KNO₃ = 39g+14g+16g×3 = 101 g/mol
1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃
:)
Answer:
Fe₂(SO₄)₃ + 6KOH —> 3K₂SO₄ + 2Fe(OH)₃
The coefficients are: 1, 6, 3, 2
Explanation:
__Fe₂(SO₄)₃ + __KOH —> __K₂SO₄ + __Fe(OH)₃
To determine the correct coefficients, we shall balance the equation. This can be obtained as follow:
Fe₂(SO₄)₃ + KOH —> K₂SO₄ + Fe(OH)₃
There are 2 atoms of Fe on the left side and 1 atom on the right side. It can be balance by writing 2 before Fe(OH)₃ as shown below:
Fe₂(SO₄)₃ + KOH —> K₂SO₄ + 2Fe(OH)₃
There are 6 atoms of OH on the right side and 1 atom on the left side. It can be balance by writing 6 before KOH as shown below:
Fe₂(SO₄)₃ + 6KOH —> K₂SO₄ + 2Fe(OH)₃
There are 6 atoms of K on the left side and 2 atoms on the right side. It can be balance by writing 3 before K₂SO₄ as shown below:
Fe₂(SO₄)₃ + 6KOH —> 3K₂SO₄ + 2Fe(OH)₃
Now, the equation is balanced.
Therefore, the coefficients are: 1, 6, 3, 2
