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2 years ago

Write the atomic symbol for the isotope of bismuth with 125 neutrons.

Chemistry

1 answer:

vlada-n [284]2 years ago

5 0

First, isotopes are the atoms of a single element whose nuclei have a different number of neutrons, and therefore, differ in mass numbers. You should know that atoms are formed by a nucleus that has a small size and is made up of protons and neutrons. The nucleus is surrounded by a cloud of electrons, which are found in a region of the atom called the cortex.

The mass number, represented as A, is the sum of the number of protons and neutrons in the nucleus. On the other hand, the atomic number (Z) is the number of protons that exist in the nucleus.

The isotopes of an element X are represented as follows,

(see first attached picture)

It should be noted that the number of neutrons of a chemical element can be calculated as the difference A-Z.

The atomic and mass numbers of bismuth with 125 neutrons are:

Z = 83

A = 83 + 125 = 208

Thus, the atomic symbol of the bismuth isotope with 125 neutrons is:

(see second attached picture)

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How do the particles in matter change when a substance becomes hotter, cools down, or changes state

Archy [21]

Colder- moves slower
warmer- moves faster
changes state:
solid to gas- warmer
gas to solid- colder
something like that

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3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago

A sample of gas initially has a volume of 2.25 L at 350 K and a pressure of 1.75 atm. What will be sample pressure if the volume

IRINA_888 [86]

Answer:

8.44 atm

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 2.25 L

Initial temperature (T₁) = 350 K

Initial pressure (P₁) = 1.75 atm

Final volume (V₂) = 1 L

Final temperature (T₂) = 750 K

Final pressure (P₂) =?

The final pressure of the gas can be obtained as illustrated below:

P₁V₁/T₁ = P₂V₂/T₂

1.75 × 2.25 / 350 = P₂ × 1 / 750

3.9375 / 350 = P₂ / 750

Cross multiply

350 × P₂ = 3.9375 × 750

350 × P₂ = 2953.125

Divide both side by 350

P₂ = 2953.125 / 350

P₂ = 8.44 atm

Thus, the final pressure of the gas is 8.44 atm.

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2 years ago

Why do your ears “pop” in an airplane?

tino4ka555 [31]

Pressure buildup in ur ear

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2 years ago

An ionic compound has the general form, xcly. which option for x and y correctly completes the formula?

igor_vitrenko [27]

Answer:
Case 1:
  X =  Any element from Group I

i)   H
ii)   Li
iii) Na
   iv) K
v)   Rb
vi)  Cs

  Y =  1

Case 2:
  X = Any element from Group II

i) Be
ii) Mg
iii) Ca
iv) Sr
v) Ba
vi) Ra

  Y = 2

Case 3:
  X =  Any element from Group III

i)   B
ii)   Al
iii) Ga
   iv) In
v)   Ti

  Y =  3

Explanation:
The general formula given is as follow,

XCly

So, if X has +1 oxidation state, then it will require only one Cl atom with oxidation number -1 to form a neutral compound, therefore, y = 1.

If X has +2 oxidation state, then it will require two Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 2.

If X has +3 oxidation state, then it will require three Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 3.

8 0

3 years ago