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iren [92.7K]
2 years ago
7

PLEASEEEEEEEEEEEEEEEEE HELPPPPPPPPPPPPPPPPPPPPPPPP

Chemistry
1 answer:
Sophie [7]2 years ago
7 0

give a very origin example a student that include B sin displayed

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Predict the missing component in the nuclear equation
Naddik [55]

The answer is C since both charge and mass have to be balanced on both sides of the equation.

7 0
1 year ago
Several properties of water are shown. Classify each property as a physical property or a chemical property.
34kurt

Chemical property:

  • Can be split into hydrogen and oxygen
  • Reacts with certain metals

Physical property:

  • Is liquid at room temperature
  • Has a density of 1.0 g/cm³
7 0
2 years ago
How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?
solong [7]

Answer:

Mass= 2.77g

Explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g

6 0
3 years ago
A student places an object on a balance. Which measurement might the student record based on the balance reading? A. 2.4 kilogra
gtnhenbr [62]
A would be the correct answer
8 0
1 year 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
juin [17]

Answer:

a)   1.12 MeV / nucleon

b)   5.62 MeV / nucleon

c)  8.80 MeV / nucleon

d) 8.56 MeV / nucleon

we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

<u>a) ²H = 1 neutron , 1 proton = 2 nucleons </u>

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm  = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3  * 931.5 ) MeV / 2 nucleons

                                                  = 1.12 MeV / nucleon

<u>b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons </u>

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 )  = 7.0582 u  

Δm  = ( 7.0582 u  - 7.0160 u  ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

                                                  = 5.62 MeV / nucleon

<u>C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons </u>

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

                    = 62.5142 u

Δm = 0.5859 u  

∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

                                                  = 8.80 MeV / nucleon

<u>D) ¹¹⁰Cd = 48 protons , 62 neutrons = 110 nucleons </u>

Theoretical mass = 109.9030 u

Actual mass = ( 48 * 1.0078 ) + ( 62 * 1.0087 )

                    = 110.9138 u

Δm  = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

                                                  = 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

3 0
3 years ago
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