Question #5ce05 - Socratic Now for the tricky part When chemists report an elemental analysis and it adds up to less than 100%, the difference is generally attributed to oxygen Our analytical methods, constrained by working in an oxygenated and water-laden world, do not in general detect or measure oxygen directly Instead we infer it by difference after accounting for the other elements
Question #88e62 - Socratic The mass of hydrogen gas is "27 g" You are asked to determine the mass of a gas in a weather balloon filled with hydrogen, therefore, the gas is hydrogen gas, "H"_2 First you need to determine the moles of hydrogen gas at STP by using the ideal gas law Once you know moles, you will determine the mass by multiplying by the moles by the molar mass of "H"_2" ("2 016 g mol") Equation for the
Question #41c83 - Socratic We assess TWO reactions (i) K(s) + H_2O(l) rarr KOH(aq) + 1 2H_2(g)uarr (ii) KOH(aq) + HCl(aq) rarr KCl(aq) + H_2O(g) And thus the moles of hydroxide is equal to the moles of metallic potassium And we can get the moles of hydroxide from the moles of hydrochloric acid "Moles of HCl"=8 00*mLxx10^-3*L*mL^-1xx1 0*mol*L^-1=8 0xx10^-3*mol And thus there were 8 0xx10^-3*mol with respect to
Question #848e1 - Socratic 12 5 Taking the reciprocal of a number means to divide 1 by it So in our case, the reciprocal is: 1 (5 12) In the case of fractions, there happens to be a really useful division rule you can use: a (b c)=a*c b So in our case, we get: 1 (5 12)=1*12 5=12 5 This means that the reciprocal of 5 12 is 12 5
Question #881a0 - Socratic h=-4 7 To eliminate the fractions in the equation, multiply ALL terms on both sides by the color (blue)"lowest common multiple" (LCM) of the denominators 4, 2 and 8 the LCM of 4, 2 and 8 is 8 rArr (cancel (8)^2xx (5h) cancel (4)^1)+ (cancel (8)^4xx1 cancel (2)^1)= (cancel (8)^1xx (3h) cancel (8)^1) rArr10h+4=3hlarrcolor (red)"no fractions" subtract 3h from both sides 10h-3h+4=cancel (3h
Question #c9e1c - Socratic OK, because the problem statement is ambiguous, I'd rather try to prove a more general and--presumably--more useful result I shall prove that, if a b is rational and x is irrational, then their sum a b + x is irrational (a and b are integers and b!=0) I'll use a proof by contradiction: let's assume that the sum of the two numbers is rational This means that the sum can be written as a ratio
Question #21b22 - Socratic Julia deposited $108 24 in her checking account This is 60% of the remaining money It follows then, that 1% of the remaining money equals "$108 24" 60 = $1 804 So all of the remainder is 100 (%) x $1 804 = $180 40 this remainder was 80% of the smount of her paycheck The money she put in her savings account was 20% of the paycheck if 80% of the paycheck is $180 40, then 10% of it is "$180