Answer:
![Na3PO4 + 3AgNO3 -------> Ag3PO4 + 3NaNO3](https://tex.z-dn.net/?f=Na3PO4%20%2B%203AgNO3%20-------%3E%20Ag3PO4%20%2B%203NaNO3)
In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M
The answer is A) PO43- < NO3- < Na+
Explanation:
Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-
We have 100mL of each reactant with the same concentration for both (1.0 M) so:
(0.1)(1)(3)= 0.3 mol Na+
(0.1)(1)= 0.1 mol NO3-
so PO43- < NO3- < Na+
Answer:
F. 2NO + 02 —> 2NO
H. 4NH3 + 502 —> 4NO + 6H20
Explanation:
The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.
2NO + 02 —> 2NO
From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.
4NH3 + 502 —> 4NO + 6H20
From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.
The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.
This is certified by the law of conservation of mass.
The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side
Answer : The number of iron atoms present in each red blood cell are, ![1.077\times 10^9](https://tex.z-dn.net/?f=1.077%5Ctimes%2010%5E9)
Explanation :
First we have to calculate the moles of iron.
![\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20iron%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20iron%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20iron%7D%7D%3D%5Cfrac%7B2.90g%7D%7B55.85g%2Fmole%7D%3D0.0519moles)
Now we have to calculate the number of iron atoms.
As, 1 mole of iron contains
number of iron atoms
So, 0.0519 mole of iron contains
number of iron atoms
Now we have to calculate the number of iron atoms are present in each red blood cell.
Number of iron atoms are present in each red blood cell = ![\frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BNumber%20of%20iron%20atoms%7D%7D%7B%5Ctext%7BTotal%20number%20of%20red%20blood%20cells%7D%7D)
Number of iron atoms are present in each red blood cell = ![\frac{3.125\times 10^{22}}{2.90\times 10^{13}}](https://tex.z-dn.net/?f=%5Cfrac%7B3.125%5Ctimes%2010%5E%7B22%7D%7D%7B2.90%5Ctimes%2010%5E%7B13%7D%7D)
Number of iron atoms are present in each red blood cell = ![1.077\times 10^9](https://tex.z-dn.net/?f=1.077%5Ctimes%2010%5E9)
Therefore, the number of iron atoms present in each red blood cell are, ![1.077\times 10^9](https://tex.z-dn.net/?f=1.077%5Ctimes%2010%5E9)
Answer:
PH= 6.767 (answer is the A option)
Explanation:
first we need to correct the value in Kw at this temperature is 2.92*10^-14
so, in this case we have that:
Kw=2.92*10^-14 M²
[ H3O^+] [ H3O^+]
![[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%5BOH%5E%7B-%7D%20%20%5D%20%3D%20Kw%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D%20%20%20%5C%5C%5C%5C)
at 40ºC
![[H_{3}O^{+} ] = [OH^{-} ]](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%5BOH%5E%7B-%7D%20%20%5D)
![[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%5E%7B2%7D%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D)
![[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%282.92%2A10%5E%7B-14%7D%29%5E%7B1%2F2%7D%20%3D%201.71%2A10%5E%7B-7%7D%20M)
![PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767](https://tex.z-dn.net/?f=PH%3D%20-log10%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20-log10%281.71%2A10%5E%7B-7%7D%20%29%20%3D%206.767)
Answer:
wait I AM TRYING..................
this is limiting reactant