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irga5000 [103]
3 years ago
13

Calculate the number of moles 41.0 g of Ca(NO3)2  1.48 g of Ca(OH)2  3.40 g of CaSO4

Chemistry
1 answer:
Lemur [1.5K]3 years ago
8 0
First. moles is just a label for a number of things. just like a dozen = 12, a gross = 144, a mole = 6022 with another 20 zeros after the 2

next

moles = mass / molecular weight.

molecular weight = sum of atomic mass from the periodic table

atomic mass MnO2 = atomic mass Mn + 2 x atomic mass O
= 54.94 + 2 x 16 = 86.94 g/mole

so moles MnO2 = 98.0 grams / (86.94 g/mole) = 1.13 moles

notice that I only gave 3 digits? that because of sig figs read the link below if you don't understand....

mw C5H12 = 5 x 12 + 12 x 1 = 72 g/mole

so moles C5H12 = 12.0 g / 72.0 g/mole = 0.167 moles

mw XeF6 = 131.3+ 6 x 19.00 = 245.3

so moles XeF6 = 100 g / 245.3 g/mole = 0.4077 moles

I've also provided a link to a periodic table. if you need atomic weights click on any element and it will give you the
details.
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2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to
evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

3 0
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if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar
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Answer:

A beaker  

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

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