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3 years ago

Select all that apply. which of the following are density labels? kg/l g/m g/ml cm^3/g

Chemistry

2 answers:

Strike441 [17]3 years ago

4 0

I would say that only the only density label is cm^3/g

valkas [14]3 years ago

4 0

The density is defined as the ratio of mass of a substance to volume of substance

density = mass / volume

the mass of any substance is measured either in grams or Kilo grams

while volume is measured as : cm^3 or mL or L or m^3

Thus correct answer will be

kg/l

g/ml

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Which of the following would be appropriate reason(s) for scientists to reject atomic theory?

leonid [27]

It posed a contradiction to Quantum Theory

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3 years ago

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Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling

Reil [10]

Answer:

The boiling point of a 8.5 m solution of Mg3(PO4)2 in water is 394.91 K.

Explanation:

The formula for molal boiling Point elevation is :

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b}$ = elevation in boiling Point

$K_{b}$ = Boiling point constant( ebullioscopic constant)

m = molality of the solution

i = Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

$Mg_{3}(PO_{4})_{2}\rightarrow 3Mg^{2+} + 2 PO_{4}^{3-}$

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

i = Van't Hoff Factor = 5

m = 8.5 m

$K_{b}$ = 0.512 °C/m

Insert the values and calculate temperature change:

$\Delta T_{b} = iK_{b}m$

$\Delta T_{b} = 5\times 0.512\times 8.5$

$\Delta T_{b} = 21.76 K$

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

$\Delta T_{b} = T_{b} - T_{b}_{pure}$

$T_{b}_{pure}$ = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

8 0

3 years ago

The smallest unit of an element that retains all the characteristics of that element is called a(n)

Maksim231197 [3]

The smallest unit of an element that retains all the characteristics of that element is called an A. atom.
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3 years ago

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Reptile [31]

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3 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

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we will use the following equation,

193x + 191(1-x) = 192.217

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60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

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3 years ago