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algol [13]
3 years ago
11

Calculate the nuclear binding energy in mega-electronvolts (MeV) per nucleon for 136 Ba . 136 Ba has a nuclear mass of 135.905 a

mu . nuclear binding energy per nucleon:
Chemistry
1 answer:
sergey [27]3 years ago
5 0

Answer:

8.194 Mev per nucleon

Explanation:

Mass of Barium = 135.905 amu

number of proton = 56, number of neutron = 80

Md = (Mp + Mn) - Mb Mp is the mass of proton, Mn is the mass of neutron, Mb is the mass of barium and Md is the mass defect

Mn = 1.00867 amu Mp = 1.00728 amu

Md = ( 56 ( 1.00728) + 80 ( 1.00867) = 137.1013 - 135.905 =1.1963 amu

Md = 1.1963 × 1 ÷ ( 6.02214 × 10 ²⁶ amu ) = 1.9865 × 10 ⁻²⁷ kg

Energy = mc² = 1.9865 × 10 ⁻²⁷ kg × (2.99792 × 10 ⁸ m/s)²

E= 1.78537 × 10⁻¹⁰ J

to convert to Mev

1.78537 × 10⁻¹⁰ × 6241457006000 = 1114.33 Mev

binding energy per nucleon = 1114.33 / 136 =8.194 Mev per nucleon

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Structure of tetrazine
Vika [28.1K]

Answer:

Tetrazine is a compound that consists of a six-membered aromatic ring containing four nitrogen atoms with the molecular formula C2H2N4.

(See the image)

Hope it helps!

3 0
3 years ago
Water flows from the bottom of a large tank where the pressure is 100 psig through a pipe to a turbine which produces 5.82 hp. T
Marianna [84]

Explanation:

Bernoulli equation for the flow between bottom of the tank and pipe exit point is as follows.

   \frac{p_{1}}{\gamma} + \frac{V^{2}_{1}}{2g} + z_{1} = \frac{p_{2}}{\gamma} + \frac{V^{2}_{2}}{2g} + z_{2} + h_{f} + h_{t}

    \frac{(100 \times 144)}{62.43} + 0 + h[tex] = [tex]\frac{(50 \times 144)}{(62.43)} + \frac{(70)^{2}}{2(32.2)} + 0 + 40 + 60

                          h = \frac{(50 \times 144)}{(62.43)} + \frac{(70)^{2}}{2(32.2)} + 40 + 60 - \frac{(100 \times 144)}{(62.43)}

                            = 60.76 ft

Hence, formula to calculate theoretical power produced by the turbine is as follows.

                                 P = mgh

                                     = 100 \times 60.76

                                     = 6076 lb.ft/s

                                     = 11.047 hp

Efficiency of the turbine will be as follows.

                \eta_{t} = \frac{P_{actual}}{P_{theoretical}} × 100%

                                = \frac{5.82}{11.047} \times 100%                      

                                = 52.684%

Thus, we can conclude that the efficiency of the turbine is 52.684%.

4 0
3 years ago
Which of the following is a single replacement reaction?
Morgarella [4.7K]

Fe + 3NaBr → FeBr3 + 3Na

The Na is replaced by the Fe atom.

4 0
3 years ago
Read 2 more answers
A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i
erastova [34]

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

7 0
3 years ago
The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:
motikmotik

The allowable combination for the atomic orbital is n=3, l=1, m_{l}=-1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; m_{l} values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Since m_{l} values are integers from -l to 0 to +l, for l = 0 the value of m_{l} cannot be -1 (l = 0 has m_{l}= 0).

There are two l values that are consistent with n and m_{l} values:

l=1 or 2

Therefore, the allowable condition is n=3, l=1, m_{l}=-1.

To know more about quantum numbers, visit: brainly.com/question/16979660

#SPJ4

4 0
2 years ago
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