Odysseus

Q1. Balance the following equation using the smallest set of yet numbers. The sum of the total coefficients is: __ SF4 + __ piddle ( __ H2SO3 + __ HF  ascendent: We intent first to start out with 1 speck sec tetrafluoride, SF4. That contains 1 sulfur, S, atom and indeed we try 1 soupcon of sulfurous acid, H2SO3, on the salutary It contains also 4 farad atoms,F, frankincense we try 4 molecules of hydrogen fluoride on the rightly Now we become 2 H atoms and 3 O atoms in H2SO3, and 4 H atoms in the 4 HF molecules on the right and that corresponds to 3 H2O molecules on the left. thence: SF4 + 3 H2O ( H2SO3 + 4 HF S: 1 on the left, 1 on the right F: 4 on the left, 4 on the right H: 6 on the left, 6 on the right O: 3 on the left, 3 on the rightOK Sum of all coefficients (do non entomb the coefficients of 1 which argon not written in the equation): 1 + 3 + 1 + 4 = 9choice A) A) 9 B) 6 C) 7 D) 4 E) 5 irregular# 3-6  place# 60 Q2. afterwards carrying out the mathematical doings below, how many strong figures are appropriate to show in the provide? (13.7 + 0.027)/8.221 issue: 13.7 + 0.027 = 13.727. We must(prenominal) have 1 significant digit later on the catch as in 13.7 (0.027 has 2), thus the first public presentation gives 13.7, which has 3 significant figures, while 8.221 has 4. Thus the final issuing must have 3 significant figures.

choice A) A) 3 B) 2 C) 4 D) 1 E) 5 Sec# 1-5 sucker# 70 Q3. Acetic acid boils at 244.2(F. What is its boiling point in degrees Celsius? Solution: With the conversion equation we get the extend: [pic] 1 To 4 significant figures as in 244! .2 that is 117.9oCchoice A) A) 117.9(C B) 167.7(C C) 153.4(C D) 382.0(C E) 103.7(C Sec# 1-3  Grade# 70 Q4. manganese makes up 1.3 x 104 percent by mass of the elements found in a normal strong body. How many grams of manganese would be found in the body of a person weighing 183 lb? (2.2 lb = 1.0 kg) Solution: 1.3 has 2 significant figures, thus the result must have...If you destiny to get a bounteous essay, order it on our website: BestEssayCheap.com

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