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Alisiya [41]
3 years ago
5

A liquid is found to have a volume of 50 mL (in a graduated cylinder). When placed on a balance, the liquid and graduated cylind

er has a mass of 125 g. The empty graduated cylinder has a mass of 75 g. What is the density of the liquid?
Chemistry
2 answers:
DanielleElmas [232]3 years ago
7 0
Heyyyyyyyyyyyyyyyyyyyyyy
aliina [53]3 years ago
5 0

Answer:

1 g/ml^3

Explanation:

Volume= 50 mL

mass= 125 initial mass-75 graduated cylinder= 50 g

d= m/v , 50g/50mL

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A. Describe the molecule chlorine dioxide, CIO in terms of three possible resonance structures.
cupoosta [38]

Answer:

See explanation

Explanation:

The compound ClO2 has 19 valence electrons.  ClO2 is a bent molecule with tetrahedral electron pair geometry but has two  lone pairs of electrons. This is indicated by the presence of four electron pairs on the outermost shell of the central atom.

The molecule has an odd number of valence electrons, hence, it is generally regarded as a paramagnetic radical. None of the proposed Lewis structures for the molecule is satisfactory because none of them obeys the octet rule.

From the images attached, one can easily see that the electron dots around the oxygen and chlorine atoms does not satisfy the octet rule in all the resonance structures shown.

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3 years ago
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butalik [34]

Answer:

C and D

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3 years ago
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Determine the number of grams in each of the quantities<br><br> 1.39.0 x 1024 molecules Cl2
STatiana [176]

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

This refers to the atomic total of the 12 gr C-12  which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31

mass Cl₂(MW=71 g/mol) :

\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01

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3 years ago
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NeTakaya
They connect at the joint
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3 years ago
How many liters of 15.0 molar NaOH stock solution will be needed to make 17.5 liters of a 1.4 molar NaOH solution? Show the work
strojnjashka [21]
2.0 L
The key to any dilution calculation is the dilution factor

The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.

In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to

DF=18.5M1.5M=12.333

So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.

The volume of the stock solution needed for this dilution will be

DF=VdilutedVstock⇒Vstock=VdilutedDF

Plug in your values to find

Vstock=25.0 L12.333=2.0 L−−−−−

The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.

So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.

IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!

In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.

Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!

So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.

Always remember

Water to concentrated acid →.NO!

Concentrated acid to water →.YES!
8 0
4 years ago
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