Fluorine reacts with hydrogen (H) and deuterium (D) to form hydrogen fluoride (HF) and deuterium fluo- ride (DF), where deuterium (21H) is an isotope of hydrogen. Would a given amount of fluorine react with different masses of the two hydrogen isotopes

The question is incomplete, the complete question is;

Fluorine reacts with hydrogen (H) and deuterium (D) to form hydrogen fluoride (HF) and deuterium fluoride (DF), where deuterium (^2_1H) is an isotope of hydrogen. Would a given amount of fluorine react with different masses of the two hydrogen isotopes? Does this violate the law of definite proportion? Explain.

No

Explanation:

The law of Definite proportions as postulated by Proust states that;

"all pure samples of a particular chemical compound contain similar elements combined in the same proportion by mass."

The two isotopes of hydrogen both represent the same element. The only difference between the isotopes is in their atomic mass. However, the both compounds HF and DF contain hydrogen and Florine atoms in exactly the same ratio by mass as postulated by the law of definite proportion.

Hence, the law of definite proportions is not violated.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.