H2SO4 is an acid
the chemical equation showing how H2SO4 is an acid according to the Arrhenius definition is as below
H2SO4 dissociate to give 2H^+ and SO4^2-
that is H2SO4 = 2H^+ + SO4^2-
According to Arrhenius an acid dissociate to give H^+ ions H2SO4 is an acid since it dissociate to give two hydrogen ions
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
The scale of most metal characteristics goes from the bottom left-hand corner.
The least metallic is the top right-hand.
So then that means that
Calcium-YES, second column
Germanium-No, to far, in the middle
Arsenic-Non-metal,
Bromine, same for this
Calcium
Answer:
D
Explanation:
the production of an odor would indicate that the heat ignited a chemical reaction
<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
