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Nina [5.8K]
3 years ago
7

What determines the order of the elements in the modern periodic table ?

Chemistry
1 answer:
Novay_Z [31]3 years ago
7 0
The last electron on the outer most shell. or d oder of dere atomic no
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The mass of a proton is 1.00728 amu andthat of a neutron is 1.00867 amu. What is the binding energy pernucleon (in J) of a Co nu
andreyandreev [35.5K]

Answer:

The binding energy per nucleon = 1.368*10^-12  (option D)

Explanation:

<u>Step 1:</u> Data given

The mass of a proton is 1.00728 amu

The mass of a neutron is 1.00867 amu

The mass of a cobalt-60 nucleus is59.9338 amu

Step 2: Calculate binding energy

The mass defect = the difference between the mass of a nucleus and the total mass of its constituent particles.

Cobalt60 has 27 protons and 33 neutrons.

The mass of 27 protons = 27*1.00728 u = 27.19656 u

The mass of 33 neutrons = 33*1.00867 u = 33.28611 u

Total mass of protons + neutrons = 27.19656 u + 33.28611 u = 60.48267 u

Mass of a cobalt60 nucleus = 59.9338 amu

Mass defect = Δm = 0.54887 u

ΔE =c²*Δm

ΔE = (3.00 *10^8 m/s)² *(0.54887 amu))*(1.00 g/ 6.02 *10^23 amu)*(1kg/1000g)

Step 3: Calculate binding energy per nucleon

ΔE = 8.21 * 10^-11 J

8.21* 10^-11 J / 59.9338 = 1.368 *10^-12

The binding energy per nucleon = 1.368*10^-12  (option D)

3 0
3 years ago
How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is ex
Brut [27]
When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M 
so by substitution we can get [Ca2+] 
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L

∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
                                                      = 0.00167 mol

4 0
3 years ago
Suppose you have 100 grams of radioactive plutonium-239 with a half-life of 24,000 years. how many grams of plutonium-239 will r
Alexandra [31]

To solve this problem, let us first calculate for the rate constant k using the half life formula:

t1/2 = ln 2 / k

where t1/2 = half life period = 24,000 years, therefore k is:

k = ln 2 / 24,000

k = 2.89 x 10^-5 / yr

 

Now we use the rate equation:

A = Ao e^(-k t)

where,

A = mass of Plutonium-239 after number of years

Ao = initial mass of Plutonium-239

t = number of years

 

A. t = 12,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 12,000)

A = 70.7 g

 

B. t = 24,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 24,000)

A = 50 g

 

C. t = 96,000 years, find A

A = 100g e^(- 2.89 x 10^-5 * 96,000)

<span>A = 6.24 g</span>

5 0
3 years ago
What is the Energy of a wave whose frequency is 8.45 x 10^14 Hz?
allsm [11]

Answer:

E = hf

E = hc/lamda

where E-energy,

h-planck's constant (6.62 x 10^-34 Js)

f- frequency ( f = c/lamda)

c-velocity of light ( 3 x 10^8 m/s)

lamda- wavelength

you can this formula to solve all the 4 questions.

8 0
3 years ago
Atomic size generally _____.
Lunna [17]
A. Increases as you move from left to right across a period
4 0
3 years ago
Read 2 more answers
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