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Rashid [163]
3 years ago
8

Suppose you are asked to find the area of a rectangle that is 2.1-cm wide by 5.6-cm long. Your calculator answer would be 11.76

cm2 . Now suppose you are asked to enter the answer to two significant figures. (Note that if you do not round your answer to two significant figures, your answer will fall outside of the grading tolerance and be graded as incorrect.)
Chemistry
1 answer:
mixas84 [53]3 years ago
8 0
1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant.2) ALL zeroes between non-zero numbers are ALWAYS significant.3) ALL zeroes which are SIMULTANEOUSLY to the right of the decimal point AND at the end of the number are ALWAYS significant.<span>4) ALL zeroes which are to the left of a written decimal point and are in a number >= 10 are ALWAYS significant.

so it will be 11.76 = 12 bc 7 rounds up

</span>
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4 0
2 years ago
If you have a gold brick that is 2cm by 3cm by 4cm and has density of 19.3g/cm3, what is its mass?
natulia [17]
Volume = a x a x a  

V = 2 cm x 3 cm x 4 cm => 24 cm³

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4 0
3 years ago
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed? 2Al + 6HCl → 2AlCl
4vir4ik [10]

Well, we need to find the ratio of Al to the other reactant.


Al:HCl = 1:3


--> this means that for every 1 Al used, you have to use 3 HCl.



6*3 = 18 moles of HCl needed to fully react with 6 moles of Al. Since 13<18, HCL is the limiting reactant.



The ratio of HCl:AlCl = 3:1



13/3 = 4.3333...



The final answer is HCl is the limiting reactant with 4.3 moles of AlCl3 able to be produced.



Hope this helps!!! :)


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