Answer :
The concentration of
before any titrant added to our starting material is 0.200 M.
The pH based on this
ion concentration is 0.698
Explanation :
First we have to calculate the concentration of
before any titrant is added to our starting material.
As we are given:
Concentration of HBr = 0.200 M
As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion
and bromide ion
.
As, 1 M of HBr dissociates to give 1 M of 
So, 0.200 M of HBr dissociates to give 0.200 M of 
Thus, the concentration of
before any titrant added to our starting material is 0.200 M.
Now we have to calculate the pH based on this
ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Thus, the pH based on this
ion concentration is 0.698
K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
Answer:
option d.......................
The properties of a compound are different than the properties of the element that forms it
<u>Answer:</u> The density of gold in
is 
<u>Explanation:</u>
Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

We are given:
Density of gold = 
Using conversion factors:
1 lb = 453.6 g
1 feet = 12 inches
1 inch = 2.54 cm
Converting given quantity into
, we get:

Hence, the density of gold in
is 