How would the following errors affect the empirical formula for the compound? a. The student ran out of time and did not do the second heating. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know? b. The student recorded the mass of the cup + sample incorrectly and started with 2.2 g of hydrated compound but used 2.0 g in the calculations. Explain how this error will affect the calculation for the number of moles of water in the hydrate? Will the final answer be artificially high or low? How do you know?

what are the óptions. pléase list them

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

the answer to your question is c

the answer to this question is 3.