Answer:
8. the answer is B.
9. the answer is A.
Explanation:
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8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.
9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:

Thus, the answer is this case is A.
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Answer:
negative but dont quote me on that
Explanation:
Answer:
=3,723.3 J=3.72 but if it has the option of -3.72 kJ then use that
Explanation:
Use the formula q=m×Cp×delta T
m=1.500 kg=1,500 g
Co=2.52 J/g·k
delta T=0.985k
q=(1,500g)(2.52 J/g·k)(0.985k)
Of Uranium-235 remains after 2.8 x 10^9 years, what was the original mass of the sample of Uranium-235? The half-life of Uranium-235 is 7.0 x 10^8 years. Uranium-232 has a half life of 68.8 years.