Plz !
what is the atomic weight of a hypothetical element consisting of two isotopes, one with mass = 64.23 amu (26.0%), and one with mass = 65.32 amu?
(a) 65.3 amu
(b) 64.4 amu
(c) 64.9 amu
(d) 65.0 amu
(e) 64.8 amu

Cryolite, na3alf6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. start this question by first balance the chemical equation.1.) balance the equation: - alo3(s)+naoh(l)+hf(> na3alf6+h2o(g). 2.) if 17.5 kilograms of al2o3(s), 51.4 kilograms of naoh(l), and 51.4 kilograms of hf(g) react completely, how many kilograms of cryolite will be produced? 3.)which reactants will be in excess, (al2o3, naoh, or hf) 4.)what is the total mass of the excess reactants left over after the reaction is complete in kg?

Imagine that a new planet is discovered with two moons of equal mass: moon a and moon b. the mass of the new planet is greater than the combined mass of its moons. moon a is farther away from the new planet than moon b. what is the planet's gravitational pull on moon a compared to the planet's gravitational pull on moon b? the planet's gravity repels moon a with a greater force than it repels moon b, which is why moon a is farther away. the gravitational pull on moon b is greater than on moon a because moon b is closer to the new planet than moon a. the gravitational pull on moon b is greater than on moon a because moon b is farther away from the new planet than moon a. the gravitational pull on moon a is the same as the gravitational pull on moon b because distance does not affect the planet's gravity.

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes