A community for students.
Here's the question you clicked on:
 0 viewing
cqluv
 one year ago
A closed container has 2.04 ⋅ 1023 atoms of a gas. Each atom of the gas weighs 1.67 ⋅ 10−24 grams. Which of the following shows and explains the approximate total mass, in grams, of all the atoms of the gas in the container?
0.34 grams, because (2.04 ⋅ 1.67) ⋅ (1023 ⋅ 10−24) = 3.4068 ⋅ 10−1
0.37 grams, because (2.04 + 1.67) ⋅ (1023 ⋅ 10−24) = 3.71 ⋅ 10−1
3.41 grams, because (2.04 ⋅ 1.67) ⋅ (1023 ⋅ 10−24) = 3.4068
3.71 grams, because (2.04 + 1.67) ⋅ (1023 ⋅ 10−24) = 3.71
cqluv
 one year ago
A closed container has 2.04 ⋅ 1023 atoms of a gas. Each atom of the gas weighs 1.67 ⋅ 10−24 grams. Which of the following shows and explains the approximate total mass, in grams, of all the atoms of the gas in the container? 0.34 grams, because (2.04 ⋅ 1.67) ⋅ (1023 ⋅ 10−24) = 3.4068 ⋅ 10−1 0.37 grams, because (2.04 + 1.67) ⋅ (1023 ⋅ 10−24) = 3.71 ⋅ 10−1 3.41 grams, because (2.04 ⋅ 1.67) ⋅ (1023 ⋅ 10−24) = 3.4068 3.71 grams, because (2.04 + 1.67) ⋅ (1023 ⋅ 10−24) = 3.71

This Question is Open

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Did you mean to write \[10^{23}\]\[10^{24}\] It can also be written as 10^23 and 10^24 Since in the problem it is written correctly (2.04 * 10^23) and (1.67 * 10^24) you can rule out the 2 solutions that have the + sign in it. Which leaves us with: (2.04*1.67) * (10^23*10^24) Without using a calculator we can round and rewrite the problem as (2)(2)(10^23)(10^24). Which is: \[4*10^{2324} = 4*10^{1} = 0.4\] If you put in the calculator (2.04*1.67) * (10^23*10^24) you'll get the exact value.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.