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Hunter-Best [27]
4 years ago
11

Which of the following atoms contains the same number of neutrons as an atom of Fluorine-19?

Chemistry
2 answers:
Pani-rosa [81]4 years ago
8 0
In Flourine-19, you can determine that the number of Neutrons is 10.

Now, if you look at Neon-20, you subtract its mass by the number of protons, leaving you with 10.

Therefore, Neon-20 has the same number of neutrons as Flourine-19.
Scorpion4ik [409]4 years ago
8 0

Answer:

Neon is the elements with same number of neutrons as of Fluorine.

Explanation:

Mass number of an element is the sum of atomic number of element and number of neutrons.

atomic number is number of protons.

The atomic number of given elements are

Neon: 10

Oxygen: 8

Sodium: 11

Nitrogen: 7

Fluorine: 9

The number of neutrons will be

Fluorine-19

Number of neutrons =19-9=10

Neon-20 :

Number of neutrons = 20-10 = 10

Oxygen-19:

Number of neutrons = 19-8 = 11

Sodium-22

Number of neutrons = 22-11 = 11

Nitrogen-16

Number of neutrons = 16-7 = 9

thus Neon is the elements with same number of neutrons as of Fluorine.

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4 0
3 years ago
A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder
Mazyrski [523]

Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

PV_1=nRT

where,

P = pressure of gas = 1 atm

V_1 = volume of gas = ?

T = temperature of gas = 270^oC=273+270=543K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K

V_1=9.12L

Now we have to calculate the volume at 24°C.

PV_2=nRT

where,

P = pressure of gas = 1 atm

V_2 = volume of gas = ?

T = temperature of gas = 24^oC=273+24=297K

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K

V_2=4.99L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 9.12 L

V_2 = final volume = 4.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (4.99-9.12)L

w=4.13L.artm=4.13\times 101.3J=418.4J

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

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