Scientists can calculate the number of neutrons by subtracting the atomic number from the atomic mass. The atomic number of an element is the number of protons which can be found in the nucleus of the element's atoms. The atomic mass of an element is the product of protons plus neutrons. The number of neutrons is not always equal to the number of protons. By subtracting the atomic number from the atomic mass, the difference will be how many neutrons are in the element's nucleus.
Answer : The pH of a 0.1 M phosphate buffer is, 6.86
Explanation : Given,
Concentration of acid = 0.1 M
Concentration of conjugate base (salt) = 0.1 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of a 0.1 M phosphate buffer is, 6.86
Qualitative chemical and enzyme tests helped Avery identify DNA as the transforming principle by conducting <span>a series of tests to find out if the transforming principle was DNA or a protein and the result is that </span>no protein was present but that DNA was present. Hope this answers your question.
Answer:
Explanation:
1. the 1/2 reaction that occurs at the cathode
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
2 the 1/2 reaction that occurs at the anode
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
2MnO2(s) + 8OH^-(aq) ----------> 2MnO4^- (aq) + 4H2O(l) +6e^-
E0 = -0.59v
3Cl2(g) +6e^- -------------> 6Cl^- (aq)
E0 = 1.39v
3Cl2 (g) + 2MnO2 (s) + 8OH^(−) (aq)---------> 6Cl^(−) (aq) + 2MnO4^(−) (aq) + 4H2O (l)
E0cell = 0.80v
Answer:
magnesium + hydrochloric acid → hydrogen gas + magnesium chloride
explanation:
the nitrogen in HNO3 is in the +5 oxidation state and is easily reduced. The reduction would result in the oxidation of the hydrogen gas, forming the water once again.The sulfur in H2SO4 is also in its highest oxidation state, +6.
<em>Hope</em><em> this</em><em> helps</em><em> </em><em>:</em><em>)</em>
