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Delicious77 [7]
3 years ago
8

Four (4.0) liters of helium gas are stored at 125 atm pressure. If the temperature and number of particles do not change, what w

ill be the new volume when the pressure is decreased to 50 atm? Show your work.
Please answer ASAP!!
Chemistry
1 answer:
seropon [69]3 years ago
5 0

Answer:

10 L

Explanation:

The only variables are pressure and volume, so we can use Boyle's Law:

p1V1 = p2V2

Data:

p1 = 125 atm; V1  = 4.0 L  

p2 =  50 atm; V2 = ?

Calculation:

125 × 4.0 = 50V2

       500 = 50 V2

        V2 = 500/50 = 10 L

The new volume will be 10 L.

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The ability of an atom in a molecule to attract electrons is best quantified by the ________.
nevsk [136]

Answer:

b. electronegativity

Explanation:

The ability of an atom in a molecule to attract electrons is best quantified by the _Electronegativity_.

Electronegativity is the ability of an atom to attract a shared pair of electron which is loosely held. Polar molecules often possess such properties. The Halogen group (group 17) is famous for such property because it has the highest electronegativity in periodic table.  

7 0
2 years ago
What is the ionic compound for FePO4?
Harlamova29_29 [7]
The  ionic compound of FePO4 is the iron III phosphate. Or you can call it ferric phosphate or ferric orthophosphate.
Just for general informations, It's normally used in steel and metal manufacturing processes, in organic farming, and many other uses.

Hope this Helps :)
8 0
3 years ago
What is the wavenumber of the radiation emitted when a hydrogen
LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted  is 0.08\times 10^{8}m^{-1}

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=\frac{hc}{\lambda}

where,

E = energy of the radiation = 1.634\times 10^{-18}J

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength of radiation = ?

Putting values in above equation, we get:

1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m

\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}

Thus wavenumber of the radiation emitted  is 0.08\times 10^{8}m^{-1}

8 0
3 years ago
How many millimeters are equal to 0.399195L
WINSTONCH [101]
399.195 millimeters=0.399195 liters
3 0
2 years ago
What would be the major product obtained from hydroboration–oxidation of the following alkenes?
zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>

In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.

See figure 1

I hope it helps!

8 0
3 years ago
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