The problem you have written you almost have it solved. Take the moles that you have calculated and multiply that by the molecular weight to get the grams.
The STP problem:
use the moles you calculated along with 1 atm for Pressure, and 273 for the temperature and plug into the PV = nRT equation. (also use 0.0821 for R)
From there you can solve for the volume
Hope this helps!
Answer:
The value of Q must be less than that of K.
Explanation:
The difference of K and Q can be understood with the help of an example as follows
A ⇄ B
In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]
In this case ratio of [B] and [A] that is
K = [B] / [A] which is called equilibrium constant.
But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.
Q = [B] / [A] . It is clear that Q < K before equilibrium.
If Q < K , reaction will proceed towards equilibrium or forward reaction will
proceed .
Answer:
16
Explanation:
4 nitrogen (N) atoms and 12 (3×4) atoms in Hydrogen (H)
Total = 4 + 12 = 16 molecules
The cell notation is:
here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell
in anodic half cell : oxidation takes place [loss of electrons]
in cathodic half cell: reduction takes place [gain of electrons]
1) this is a galvanic cell
2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode
Therefore
3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative
Δ
As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.
Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
