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Anna007 [38]
3 years ago
5

How many grams of NaCI are present in 11.00 moles

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer: 642.93 g of NaCl

Explanation:

11.00 mol NaCl  x  __58.448 g__  =  642.928 g of NaCl

                                 1 mol NaCl

I would round to 642.93 g of NaCl, but round to however many significant figures asked for.

My chemistry teacher made this map (image attached) to teach us how to do conversions. Just follow the map. I think it's pretty straight forward. I hope this helps.

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PLEASE HELP!
pogonyaev

Answer:

Q = 2.60 • 10^{3} J

Explanation:

Our specific heat capacity equation is:

Q = mC∆T

Q is the energy in joules.

m is the mass of the substance.

∆T is the temperature chance.

Let's plug in what we know.

  • We have 76.0 g of octane
  • The specific heat capacity of octane is 2.22 J/(g•K)
  • The temperature increases from 10.6º to 26.0º (a 15.4º increase)

Q = 76.0(2.22)(15.4)

Multiply.

Q = 2598.288

We want three significant figures.

Q = 2.60 • 10^{3}

or

Q = 2590 J

Hope this helps!

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2 years ago
Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer
bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0
3 years ago
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