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Viktor [21]
3 years ago
11

A container has a volume of 2.79 L and a pressure of 5.97 atm. If the pressure changes to 1460 mm Hg, what is the container’s ne

w volume?
Chemistry
1 answer:
maw [93]3 years ago
5 0

Answer:

8.68 L is the new volume

Explanation:

You use Boyle's law for this.

P_{1}V_{1}=P_{2}V_{2}

P_{1} = first pressure

P_{2} = second pressure

V_{1} = first volume

V_{2} = second volume

Convert pressure from atm to mmHg (use same units):

5.97 x 760 = 4537.2 -> 4.54 x 10³

...maintain 3 significant figures in calculation, and round as needed...

(4.54 x 10³ mmHg)(2.79 L) = (1460 mmHg)(V_{2})

(4.54 x 10³ mmHg)(2.79 L) / (1460 mmHg) = V_{2} = 8.68 L

Hope this helps :)

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Lunna [17]

Answer:

Use a ratio of 0.44 mol lactate to 1 mol of lactic acid  

Explanation:

John could prepare a lactate buffer.

He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.

\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}

He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.

For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.

6 0
3 years ago
How many atoms of oxygen (O) are in one formula unit of Mg(NO3)2?
miv72 [106K]
We may use the molecular formula of the compound to determine the number of oxygen atoms in one formula unit. The formula unit is:
Mg(NO₃)₂
Here, we can see that there are two nitrate ions in each mole, and each mole of nitrate ion contains three oxygen atoms. Thus, there are 6 oxygen atoms in each formula unit of magnesium nitrate.
5 0
3 years ago
Read 2 more answers
given that the speed of light is 2.998 x 108 m/s, what is the frequency of a wave that has a wavelength of 3.55 x 10-8 meters
Vesna [10]
Frequency, f = v / λ

f = 2.998 * 10⁸ / 3.55*10⁻⁸

f = 8.445 * 10¹⁵ Hz.
8 0
3 years ago
A plot of binding energy per nucleon Eb A versus the mass number (A) shows that nuclei with a small mass number have a small bin
Illusion [34]

Answer:

6He =   4.90 MeV/Nucleon

8Li =     5.18  MeV / Nucleon

62Ni =   8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its  protons and neutrons, this mass is greater than  the actual  value . This is the mass defect.

Now this mass defect we can convert to energy  by utilizing Einstein´s equation, E = mc².This is  the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons =   2 x 1.0078 u  =  2.0516 u

mass neutrons = 4 x 1.0087 u  =  4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹²  J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

8 0
3 years ago
A hydrogen atom requires a minimum energy of 2.18×10^-18 J atom to remove an electron from its ground state level. Determine whe
frosja888 [35]
Just find the energy of the <span>blueviolet light with a wavelength of 434.0 nm using the formula:

E  = hc / lambda

E = energy
c= speed of light =  3 x 10^8 m/s
h = planck's constant =  6.6 x 10^{-34}  m^2 kg / s
lambda =  434 nm =  434 x 10^{-9} m

Putting these values (with appropriate units) in the above formula :

we get:  Energy, E = 4.5 x 10^{-19} J

E = 0.45 x 10^{-18} J

Now, the </span>minimum energy is 2.18×10^-{18} J but our energy is 0.45 x 10^{-18} J which is less.
<span>Means the electron will not be removed

</span>
5 0
3 years ago
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