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Chemistry 6 Online
OpenStudy (anonymous):

okay so I have to figure out the number of M&M's in a box with using the mass of an empty box, and the mass of a box with M&M's. Data so far: mass of 1 plain M&M: .8 mass of empty box: 34.2 Box w/ M&M's: 129.2 Total mass of plain M&M's in the box: 95 mass of peanut M&M: 2.4 mass of empty box: 34.2 box w/ peanut M&M's: 127.1 Total mass of peanut M&M in box: 92.8 & the questions I need help on are: Calculate the number of plain M&M's in the box. & Calculate the number of peanut M&M's in the box.

OpenStudy (frostbite):

@brieculver Hello there. Shall we solve your prob together? This is fairly simply, but I like you to do the calculations.

OpenStudy (anonymous):

Thats fine just tell me how @Frostbite

OpenStudy (frostbite):

Well I help you with first one then you take the last one. The plain M&M problem: If we know the total mass of M&M in the box and we know the mass of 1 we can simply divide them with each other and get \[\Large n(M\&M)=\frac{ 95 }{ 0.8 }=118.75\]

OpenStudy (anonymous):

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OpenStudy (anonymous):

but wait no I need to find out how many M&M's are in the box with this data.. @Frostbite

OpenStudy (frostbite):

Exactly. That is indeed what you do. you can put units on if you like and see it is true. Lets just for fun assume the mass is in gram: \[\Large \frac{ 95 ~ gram }{ 0.8 \frac{ gram }{ M\&M } } = 118.75 ~ M\&M\] You see?

OpenStudy (frostbite):

The total mass of M&M inside the box you get from simple withdraw the mass of the box from the mass of box /w M&Ms, which they for some reason already gave you...

OpenStudy (anonymous):

oh okay I see but it didn't seem like that much. But then again looks can be perceiving but thank you!!

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