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Oliga [24]
3 years ago
13

A sample of gas (24.2 g) initially at 4.00 atm was compressed from 8.00 l to 2.00 l at constant temperature. after the compressi

on, the gas pressure was __________ atm
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
5 0
Answer: 16 atm  
Explanation: 
P1V1 = P2V2 
P2 = P1V1/V2 
=4 atm x 8.00 L/2.00L = 16 atm
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Manganese forms several oxides when combined with oxygen. One of the oxides (Oxide 1) contains 63.2% of Mn and another oxide (Ox
Nina [5.8K]

Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

3 0
3 years ago
Which of the following gases is most closely linked to day-to-day weather changes?
Olin [163]
Short answer: nitrogen, oxygen, Argon, and

inert gas

The atmosphere contains many gases, most in small amounts, including some pollutants and greenhouse gases.

These contribute to climate change.
8 0
3 years ago
How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
lisov135 [29]
Moles of glucose = Molarity x volume solution 
                              = 4.5 x 1.5
                              = 6.75 moles.

Hope this helps, have a great day ahead!
3 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
Solutions and colloids which contain similarly sized particles can be effectively separated using paper filtration.
mojhsa [17]
False cause there particles arent large enough to be filtered
6 0
3 years ago
Read 2 more answers
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