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3)  One mole = 6.0221415 x 10^23 molecules

     23 x [ 6.0221415 x 10^23] =1.385092545 x 10^24 - molecules in 23 moles of Sodium.

8)

Convert from grams into liters:

46 g of Cl_2 (chlorine)

Convert from grams to cubic centimeters using the density 0.003214 g/cm^3 (at STP) of Cl_2:

46 g (1/(0.003214 g/cm^3)) = 14310 cm^3

Convert to milliliters using the relation 1 cm^3 = 1 mL:

14310 cm^3 ((1 mL)/(1 cm^3)) = 14310 mL

Convert to liters using the relation 1000 mL = 1 L:

14310 mL ((1 L)/(1000 mL)) = 14.31 L

7) Sorry, I don't know this one!.

Thank you!!! I rlly needed help! 

1) The molar mass of HCl =36.461 g/mol (grams per mole)

20.2 / 36.461 =0.554 - moles in 20.2 g of HCl.

2) 2 x 36.461 =72.922 - grams in 2 moles of HCl.

5+4+3 =12 - total number of marbles

5/12 =41.67%

Let the number of extra green marbles =G

[G+5] / [3+4+5 + G] =0.75

Solve for G:

(G + 5)/(G + 12) = 0.75

0.75 = 3/4:

(G + 5)/(G + 12) = 3/4

Cross multiply:

4 (G + 5) = 3 (G + 12)

Expand out terms of the left hand side:

4 G + 20 = 3 (G + 12)

Expand out terms of the right hand side:

4 G + 20 = 3 G + 36

Subtract 3 G + 20 from both sides:

G = 16 - Extra green marbles will be needed to make 75% of the total.

Proof: 16 + 5 =21 - Green marbles

3 + 4 + 21 = 28 - marbles in total, so that:

21 / 28 =75% - are green marbles.

One mole = 6.0221415 x 10^23 molecules

[8.022 x 10^26] / [6.0221415 x 10^23]

=1,332.084276 - moles of NaCl.

Thank You All!!

Nothing shows up....but here's the graph :

https://www.desmos.com/calculator/gd8wrobuuc

The parent, y  =  1/x, is just shfted up 2 units

Here is the graph of the "parent"  y  = 1/x   and the given function

https://www.desmos.com/calculator/mf7f467609

The parent is  reflected about the x axis and vertically "stretched" by a factor of 2

nice job you did good

No pic shows up  .......

Don't know about the big brain part....but.....we have this exponential decay form :

22.1875   =  355 (1/2)^(t / 3.75)       we want to find "t"

Divide both sides by 355

22.1875 / 355  =  (1/2)^(t/3.75)       take the log of both sides

log  ( 22.1875 / 355)  =  log (1/2)^(t / 3.75)      and we can write

log  ( 22.1875 / 355)  =   (t / 3.75) * log (1/2)   divide both sides by log (1/2)

log  ( 22.1875 / 355) / log (1/2)  =  t / 3.75      multiply both sides by 3.75

3.75 log  ( 22.1875 / 355) / log (1/2)  =  t  =   15  years old

Note that since a half-life lasts 3.75 years....it has gone through

15  / 3.75  =   4 half-lives

That worked out kinda' nice and neat  .....!!!

Just a  slight mis-interpretation......no big deal

So we have

³√16 * 7³ √2   =

³√16  * 343√2  =

³√8  * ³√2  *  343√2  =

2 * ³√2  *  343√2  =

2 * 343  *  (2)^1/3  * (2)^1/2   =

686 * (2)^2/6 * (2)^3/6   =

686 * (2)^5/6  

totally it does thank you!!

y  =  1/x     y  =  3/x

Note that  if x  =  1   

y   in the first function  = 1/1  = 1

But y in the second function becomes  3/ 1  =  3

So....the second function vertically "stretches" the  first function by a factor of 3

And the  "-4"  at the end   will shift this y = 3/x   down 4 units 

So.....the third answer is correct

Note, NSS....we have a vertical asymptote at x  =  3

So....we cannot put 3 in the denominator of the rational function because this makes the denominator = 0.....so....the last two answers are likely candidates

Note that   y  = 2/ ( x - 3)  looks like this : https://www.desmos.com/calculator/mhbrjflsb8

But....our function is shifted down....so... the "-6"  as the second term will shift this down 6 units

So.....the third answer is correct

Does that make sense??

Thanks Hectitar but I think the correct answer is

686 x 2^5/6

I checked it with Wolfram Alpha

(http://www.wolframalpha.com/input/?i=%5E3%E2%88%9A16+x+7%5E3%E2%88%9A2)

Plus I checked it with another calculator and the answer was: 686 x 2^5/6

A liitle tricky, but we want to solve this

Total cost / Number  of Students  < Cost Per Student

275  / N   <   15        Multiply both sides by N

275 < 15N       Divide both sides by 15

18.333 < N   which means the same as

N >  18.3333......so   we need   19  students

\(\quad\sqrt[3]{16}\,\times\,7\sqrt[3]{2} \\ =\\ \quad \sqrt[3]{16}\,\times\,7\,\times\,\sqrt[3]{2} \\ =\\ \quad \sqrt[3]{16}\,\times\,\sqrt[3]{2}\,\times\,7 \\ =\\ \quad \sqrt[3]{16\,\times\,2}\,\times\,7 \\ =\\ \quad \sqrt[3]{2\,\times\,2\,\times\,2\,\times2\,\times\,2}\,\times\,7 \\ =\\ \quad \sqrt[3]{2^3\,\times\,2^2}\,\times\,7 \\ =\\ \quad \sqrt[3]{2^3}\,\times\,\sqrt[3]{2^2}\,\times\,7 \\ =\\ \quad 2\,\times\,\sqrt[3]{4}\,\times\,7 \\ =\\ \quad 2\,\times\,7\,\times\,\sqrt[3]{4} \\ =\\ \quad 14\,\times\,\sqrt[3]{4} \\ =\\ \quad 14\sqrt[3]{4} \\ \)

thanks, CPhill I owe you one

Remember, DC, that

log a  = b    in exponential form  is  10^b  = a

So

 Log_x (1 / 8) = - 3 / 2   =

x^(-3/2)  =  1/8       take each side to the  (-2/3) power

x  =  (1/8)^(-2/3)    =   [ (1/8)^(1/3) ]^(-2)   =   ( 1 /  ³√8  )^(-2)   =   (1/2)^(-2)

Note now that    (1/a)^(-m)  =  a^m

So

(1/2)^(-2)  =  (2)^2   =  4  =  x

Solve for x:

-log(8)/log(x) = -3/2

Take the reciprocal of both sides:

-log(x)/log(8) = -2/3

Multiply both sides by -log(8):

log(x) = (2 log(8))/3

(2 log(8))/3 = log(8^(2/3)) = log(4):

log(x) = log(4)

Cancel logarithms by taking exp of both sides:

x = 4

Hectictat figured this out.....I think she gave Manuel the correct answer by private message

But.....for anyone interested...it was supposed to be  :

√ 5  ,  ³√4

And we can write

5^(1/2) ,   4^(1/3)    =      {get a comon denominator of 6 on the fractions }

5^(3/6),  4^(2/6)  =

⁶√ [ 5^3 ] , ⁶√ [4^2 ]    =

⁶√125,  ⁶√16

Thanks, Rosala....it seems that Melody has helped you more recently than I have....she's WAY better at these counting and probability questions than I am....!!

I must  also give a SHOUT OUT  to hectictar and X².......their answers are always well explained

Hey....that cake looks good....!!!

Thank You

no problem