Find protons, electrons neutrons for nitrogen -3 with 7 neutrons

Write a nuclear equation to describe the neutron induced fission of Pu-239 to form Kr-89 and Ce-149 (this fission is induced by

zlopas [31]

Answer:

$\rm 2_{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, 3_{0}^{1}\text{n}; three$

Explanation:

If the fission is induced by two neutrons, the unbalanced equation is

$2\rm _{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, x_{0}^{1}\text{n}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation .

Balancing the superscripts, we get

2 + 239 = 89 + 149 + x

241 = 238 + x

x = 3

The balanced nuclear equation is

$\rm 2_{0}^{1}\text{n} + \, _{94}^{239}\text{Pu} \longrightarrow_{36}^{89}\text{Kr} + \, _{58}^{149}\text{Ce} + \, \mathbf{3}_{0}^{1}\text{n}$

$\text{The fission produces } \boxed{\textbf{three}} \text{ neutrons}$

4 0

3 years ago

Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr

kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm 5.1 atm 0 Initial

-2x -x +2x Reacts (stoichiometry is 2:1:2)

4.9-2x 5.1-x 2x Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0

2 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a

Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL = 0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

⇒ with V1 = volume of flask with stopcock valve closed

⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened

⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

⇒ with V1 = volume of helium = 0.265 L

⇒ with P2 = The new partial pressure of helium

⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

⇒ with V1 = volume of argon = 0.465 L

⇒ with P2 = The new partial pressure of argon

⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

⇒ P(Ar) = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

5 0

3 years ago

Aonde a civilização egípcia se desenvolvera

Lorico [155]

A civilização egípcia se desenvolveu ao longo do rio Nilo em grande parte porque as enchentes anuais do rio garantiam um solo rico e confiável para o cultivo.

3 0

3 years ago