Both D and G options. Weak base and its conjugate acid or weak acid and its conjugate base are the possible components of a buffer solution.
Explanation:
Buffer solution is the solution which gets easily dissolved in water and so called as "Aqueous solution".
Buffer solution is essentially made up of two components Known as:
i.) Weak base and its conjugate acid
ii.) Weak acid and its conjugate base
This weak acid and base solution is used to maintain the pH value of the solution in a balanced way.
When the weak acid or base solution is added to strong acid or base solution that is the way pH gets balanced .
In one word buffer solution is the solution which resists for the pH change when strong acids or bases are added.
When pure HA is added to the buffer, the buffer component ratio and the pH decrease.
<h3>State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA] for the dissolve of pure HA in the buffer.</h3>
When pure HA is added to the buffer, the buffer component ratio and the pH decrease. The added HA increases the concentrations of NA and HA. However, there is a greater relative increase in the concentration of HA. Hence, the ratio of [NaA]/[HA] decreases, causing the solution to become more acidic.
The capacity of a buffer to withstand pH change is measured. The concentration of the buffer's components namely, the acid and its conjugate base determine this ability. Greater buffer capacity is associated with higher buffer concentration.
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Answer:
The strong electric force.
Explanation:
Trust me. If its wrong, comment on this, don't delete it. If someone else answers with the wrong question than other students might get the answer wrong. So if its right, say that it is, if its wrong, tell others what the real answer is.
Answer:
15.95
Explanation:
This question is a modification of the calculation of the empirical formula of a compound given its percent composition and atomic weights of the elements in the compound.
Here we are given the formula and the percent composition, so we know that there are 4 atoms of E per 2 atoms of N so lets solve using the information given.
In 100 grams of the binary compound we have
30.46 g N
69.54 g E
The number of moles is the mass divided by atomic weight:
mol N = 30.46 g / A.W N = 30.46 g / 14.00 g/mol = 2.18 mol N
mol E = 65.54 g / A.W E
Thus,
4 mol E/ 2 mol N = ( 69.54 g/ A.W E ) / 2.18
2 A.E = 65.54 g / 2.18 ⇒ A.W E = 69.54 g / ( 2 x 2.18 ) = 15.94 g
So the A.W is 15.94 g/mol which is close the atomic weight of O.
Answer: There is no question, but we can calculate a couple of items:
Density of sea water sample = (52.987g-44.317g)/8.5ml
Inorganic content of sample (mostly salts) = (44.599g-44.317g)/(52.987g-44.317g) x 100% = percent inorganics in water sample
Explanation:
