Answer:
Oxidizing agent - CrO4^2-
Reducing agent- N2O
Explanation:
Let us look at the equation closely;
CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]
The reduction half equation is;
CrO4^2- (aq) + 3e -------->Cr^3+ (aq)
Oxidation half equation is;
3N2O(g) ------>3 NO(g) +3 e
Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.
The question is incomplete, complete question is;
A solution of
is added dropwise to a solution that contains
of
and
and
.
What concentration of
is need to initiate precipitation? Neglect any volume changes during the addition.
value 
value 
What concentration of
is need to initiate precipitation of the first ion.
Answer:
Cadmium carbonate will precipitate out first.
Concentration of
is need to initiate precipitation of the cadmium (II) ion is
.
Explanation:
1) 
The expression of an solubility product of iron(II) carbonate :
![K_{sp}=[Fe^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=2.10%5Ctimes%2010%5E%7B-11%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B2.10%5Ctimes%2010%5E%7B-11%7D%7D%7B1.15%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=1.826\times 10^{-9}M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D1.826%5Ctimes%2010%5E%7B-9%7DM)
2) 
The expression of an solubility product of cadmium(II) carbonate :
![K_{sp}=[Cd^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
![1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]](https://tex.z-dn.net/?f=1.80%5Ctimes%2010%5E%7B-14%7D%3D0.58%5Ctimes%2010%5E%7B-2%7D%20M%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
![[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D%5Cfrac%7B1.80%5Ctimes%2010%5E%7B-14%7D%7D%7B0.58%5Ctimes%2010%5E%7B-2%7D%20M%7D)
![[CO_3^{2-}]=3.103\times 10^{-12} M](https://tex.z-dn.net/?f=%5BCO_3%5E%7B2-%7D%5D%3D3.103%5Ctimes%2010%5E%7B-12%7D%20M)
On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.
So, cadmium carbonate will precipitate out first.
And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the
concentration.
1. a 2.b 3 a 4. c. 4.a 5. d 6.a. 2Na + Cl2 2NaC 7.c. 3 8.3 9.b
<h2>The required option d) "specific heat" is correct.</h2>
Explanation:
- To raise the temperature of any substance or material of certain mass to respective temperature it requires some amount of heat.
- Specific heat is the amount of heat necessary to raise the temperature of the substance of 1 gram to 1 Kelvin.
- It is the amount of heat which is required to raise the temperature per unit mass to per unit temperature.
- Thus, the required "option d) specific heat" is correct.
Answer:
Check the explanation
Explanation:
AT = A0 e(-T/H)
... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.
Substituting what we know, we get...
0.71 = (1) e(-T/5730)
Solve for T...
loge(0.71) = -T/5730
T = -loge(0.71)(5730)
T = 1962 (conservatively rounded, T = 2000)
similarly for all
for aboriginal charcoal
0.28 = (1) e(-T/5730)
Solve for T...
loge(0.28) = -T/5730
T = -loge(0.28)(5730)
T = 7294 (conservatively rounded, T = 7000)
for mayan headdress
0.89 = (1) e(-T/5730)
Solve for T...
loge(0.89) = -T/5730
T = -loge(0.89)(5730)
T = 667 (conservatively rounded, T = 700)
for neanderthal
0.05 = (1) e(-T/5730)
Solve for T...
loge(0.05) = -T/5730
T = -loge(0.05)(5730)
T = 17165 (conservatively rounded, T = 17000)