In a chemical process, the amount of a certain type of impurity in the output is difficult to control and is thus a random variable. it is known that the standard deviation of the impurity is 0.0015 following an approximately normal distribution. if a random sample of 75 outputs has an average impurity of 0.310, find a 95% confidence interval for the mean amount of the impurity.

+/- 0.00033

Explanation:

For a 95 % confidence interval the range is given by

+/- Z * s/sqrt(n)

where Z value is 1.916 for a 95% confidence interval

Therefore the interval is

s is the standard deviation in this case 0.0015

n is 75 the number of outposts

calculating,

+/- 1.916 * 0.0015/sqrt(75) = +/- 0.00033

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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