This is a neutron induced fission, therefore a neutron will be added to the U²³⁵ to cause the reaction, and thus it will be added to the left side. There will be unknown number of neutrons produced and thus we put this on the right hand side.
n₁ + U²³⁵ = Te¹³⁷ + Zr ⁹⁷ + xn1 ( n1 to mean a neutron of mass 1)
To balance the masses on both sides of the equation;
1 + 235 = 137 +97+ x
x = 2
the end reaction will be
n₁ +U²³⁵ = Te¹³⁷ + Zr⁹⁷ + 2 n₁
The solution would be like this for this specific problem:
<span>Given:
H2 = </span><span>2.6 atm
CL2 = 3.14 atm</span>
<span>
pressure H2 = 2.6 - x
pressure Cl2 = 3.14 - x
<span>pressure HBr = 2x = 1.13
x = 1.13 / 2 = 0.565
<span>pressure H2 = 2.6 - 0.565 = 2.035
pressure Br2 = 3.14 - 0.565 = 2.575
Kp = (1.13)^2 / 2.035 x 2.575</span></span></span>
= 1.2769 / (5.240125)
= 0.24367739319195629875241525726963
= 0.244
<span>Therefore, the Kp for the reaction at the given temperature
is 0.244.
To add, </span>the hypothetical pressure of a gas if
it alone occupied the whole volume of the original mixture at the same
temperature is called the partial pressure or Kp.
Answer:
35.5g of sugar
Explanation:
To solve this question we must assume the density of coke = Density of water = 1g/mL
Thus, in a single can of coke, the mass is 355g. Now, the 10% of this coke is sugar. That means the amount of sugar you are consuming is:
355g * (10/100) = 35.5g of sugar
<em>10/100 = 10%</em>
The wt% of KOH = 45%
This implies that there is 45 g of KOH in 100 g of the solution
Density of the solution is given as 1.45 g/ml
Therefore, the volume corresponding to 100 g of the solution is
= 100 g * 1 ml /1.45 g = 68.97 ml = 0.069 L
Now concentration of the concentrated KOH solution is:
Molarity = moles of KOH/vol of solution
= (45 g/56.105 g.mol-1)/0.069 L = 11.6 M
Thus,
Initial KOH concentration M1 = 11.6 M
Initial volume = V1
Final concentration M2 = 1.20 M
Final volume V2 = 250 ml
M1*V1= M2*V2
V1 = M2*V2/M1 = 1.20*250/11.6 = 25.9 ml = 26 ml
The open spaces in water's crystal structure make it possible for <span>aquatic life to exist at the north pole.
It is good to know that, at the north pole, when the temperature drops below zero, only the top layer of the water freezes. Underneath this freezing layer, water remains in its liquid form, thus, allowing marine life to exist. As for oxygen, it is also trapped underneath the ice layer. </span>
