Answer:
i) pH = 2
pH = -log(H+)
:- (H+) = 10^(-2)
:- (H+) = 0.01 M
ii) pH = 6
pH = -log(H+)
:- (H+) = 10^(-6)
:- (H+) = 0.000001 M
Explanation:
By definition: pH = -log(H+).
Given your pH, solve for the H+ using the the following log rule:
if a = (+/-) log (b) then
b = 10^((+/-) a).
Also remember unit of concentration is molar (M)
Answer:
by statistical analyses, especially by determining the p-value
Explanation:
In general, observations and results obtained from experimental procedures are subjected to a statistical test to check the robustness of the working hypothesis. The p-value is the most widely used statistical index in order to test such observations and results. The p-value is the statistical probability of obtaining extreme observed results when the null hypothesis is considered correct. A p-value lesser than 0.05 generally is considered statistically significant and then the null hypothesis can be rejected. In consequence, a very low p-value (which is obtained by statistical analysis of the observations and results), indicates that there is strong evidence in support of the alternative hypothesis.
Answer:
chuma ya ndoshi.............
Explanation:
.
kwenda ukalale .............
Answer:the answer is 18.01528
Explanation:
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.
