Molarity = moles / liter
a) M = 2/4 = 0.5 M
b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M
c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M
Answer:
k = 1.3 x 10⁻³ s⁻¹
Explanation:
For a first order reaction the integrated rate law is
Ln [A]t/[A]₀ = - kt
where [A] are the concentrations of acetaldehyde in this case, t is the time and k is the rate constant.
We are given the half life for the concentration of acetaldehyde to fall to one half its original value, thus
Ln [A]t/[A]₀ = Ln 1/2[A]₀/[A]₀= Ln 1/2 = - kt
- 0.693 = - k(530s) ⇒ k = 1.3 x 10⁻³ s⁻¹
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)
Yes if you add an energy to an electron the electron will become excited, and it will jump to its highest level then go back down releasing energy
That would be the first option Ca(OH)2 + H2SO4 → CaSO4 + 2H2O.
The Ca replaces the H2 in H2SO4, and the H2 replaces the Ca is Ca(OH)2.
