Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!
http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm
<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V
<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)
</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)
<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V
V= - 0.777V (no spontaneous)
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
Answer:
large, released
Explanation:
As we know, fireworks contain l o t s of energy, even before the burst of colors release. So i think choice 3 is the answer.
i hope this helps :)
Answer is: Ksp for calcium sulfate is 2.36·10⁻⁴.
Balanced chemical reaction (dissociation):
CaSO₄(s) → Ba²⁺(aq) + SO₄²⁻(aq).
m(CaSO₄) = 0.209 g.
n(CaSO₄) = m(CaSO₄) ÷ M(CaSO₄).
n(CaSO₄) = 0.209 g ÷ 136.14 g/mol.
n(CaSO₄) = 0.00153 mol.
s(CaSO₄) = n(CaSO₄) ÷ V(CaSO₄).
s(CaSO₄) = 0.00153 mol ÷ 0.1 L = 0.0153 M.
Ksp = [Ca²⁺] · [SO₄²⁻].
[Ca²⁺] = [SO₄²⁻] = s(CaSO₄).
Ksp = (0.0153 M)² = 2.36·10⁻⁴.
Explanation:
Depression in Freezing point
= Kf × i × m
where m is molality , i is Van't Hoff factor, m = molality
Since molality and Kf remain the same
depression in freezing point is proportional to i
i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2
i=1 for C2h6O
i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)
So the freezing point depression is highest for MgCl2 and lowest for C2H6O
so freezing point of the solution = freezing point of pure solvent- freezing point depression
since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point
