Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
Answer:
Heat transfer in step 2 = 47.75 J
Explanation:
Internal energy = heat + work done
U = Q + W
In a cyclic process the total internal energy change of the system = 0.
In the process there are two steps. The total heat exchange in the process is the sum of heat exchanges in the two processes.
We have to find the heat exchange in step 2.
In step 1,
W = 1.25 J Q = -37 J
= -37 + 1.25 = -35.75 J
In step 2, the internal energy change will be negative of that in step 1.
U = 35.75 J
W = -12 J
U = Q + W
35.75 = Q -12
Q = 47.75 J
Heat transfer in step 2 = 47.75 J
Answer:-
Alpha decay
Explanation:-
Uranium 238 has atomic number 92 and mass number 238.
Thorium 234 has atomic number 90 and mass number 234.
So, the change in atomic number as uranium 238 disintegrates into thorium234 = 92 – 90 = 2
So, the change in mass number as uranium 238 disintegrates into thorium234= 238 – 234 = 4
We know that when an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.
So when an atom of uranium 238 undergoes radioactive decay to form an atom of thorium-234, alpha decay has occurred.
From the reaction between Cu and HNO₃, the formed gas is NO₂ instead of NO₃. Hence the correct balanced equation would be,
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O<span>(l)
Here, Cu goes to </span>Cu(NO₃)₂ by changing its oxidation number from 0 to +2 while NO₃⁻ goes to NO₂ by reducing its oxidation state from +5 to +4 . Hence Cu is oxidized by HNO₃ in the reaction.