Answer:
HCl
Explanation:
A limiting reactant is the lowest amount of reactant so that the reaction will stop after all that reactant used.
To answer this question, you have to change all reactant units into moles. You have 52g HCl and its molecular mass is 36.46g/mole. The number of HCl in moles will be: 52g/ (36.46g/mole)= 1.43 moles.
In the balanced reaction formula, you can see that you need one mole of HCl and one mole of NaOH for each reaction. Both molecule's coefficient is 1.
If we used up all NaOH, then the number of HCl left will be:
(moles of NaOH used * NaOH coefficient) - (moles of HCl used * HCl coefficient)
(moles of HCl you have * HCl coefficient) - (moles of NaOH used * NaOH coefficient)
(1.43 moles *1 ) - (2.5 moles*1 ) = -1.07 mole
Since the result is minus, then it means we need more HCl and we can't use all NaOH.
If we used up all the HCl, then the number of NaOH left will be:
(moles of NaOH you have* NaOH coefficient) - (moles of HCl used * HCl coefficient)
(2.5 moles*1 )- (1.43 moles *1 )= 1.07 moles
Since the result is plus, then it means we can use all HCl. Then HCl is the limiting reactant
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
Carbon has 4 electrons in its outermost shell.it shows tetravalency and catination property hence forms long chain of carbon .
Answer : The standard cell potential of the reaction is, -1.46 V
Explanation :
The given balanced cell reaction is,
Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
The standard values of cell potentials are:
Standard reduction potential of lead ![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
Standard reduction potential of chromium ![E^0_{[Cr^{3+}/Cr]}=1.33V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D%3D1.33V)
Now we have to calculate the standard cell potential for the following reaction.
![E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Therefore, the standard cell potential of the reaction is, -1.46 V
