Answer: The concentration of 
ions in vinegar is 0.001 M.
Explanation:
Given: pH = 3.0
pH is the negative logarithm of concentration of hydrogen ion.
The expression for pH is as follows.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the value into above expression as follows.
![pH = - log [H^{+}]\\3.0 = - log [H^{+}]\\conc. of H^{+} = antilog (- 3.0)\\= 0.001 M](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5C3.0%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%20antilog%20%28-%203.0%29%5C%5C%3D%200.001%20M)
Thus, we can conclude that the concentration of ions in vinegar is 0.001 M.
Answer:
D. 6.00 L
Explanation:
What we have here is an example of Boyle's Law. The equation here is P₁ · V₁ = P₂ · V₂. We know all of the values except for V₂.
60(8) = 80V
<em>Multiply 60 by 8 to get 480.</em>
480 = 80V
<em>Divide both sides by 80.</em>
480/80 = V
6 = V
The final volume for the gas is 6.00 L.
The answer is, All of the above
2NH4ClO4 --------> N2 + Cl2 + 2O2 + 4H2O
from reaction 2 mol 1 mol
given x mol 0.10 mol
Proportion:
<u>2 mol NH4ClO4 </u>= <u>1 mol Cl2</u>
x mol NH4ClO4 0.10 mol Cl2
x= (2*0.10)/1 = 0.20 mol NH4ClO4
