Which front is a head on collision or neither warm nor cold air moves
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Answer : The change in enthalpy of the reaction is, -310 kJ
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given main reaction is,
The intermediate balanced chemical reaction will be,
(1)
(2)
(3)
Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :
(1)
(2)
(3)
The expression for enthalpy of formation of will be,
Therefore, the change in enthalpy of the reaction is, -310 kJ
Answer:
a) E = k Ze (1- r³ / R³) 1/r², b) E=0, c) E = -6.62 10¹⁰ N / C
Explanation:
a) For this we can use the law of Gauus
Ф = E- dA = / ε₀
where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product
E A =q_{int} / ε₀
the area of a sphere
A = 4π r²
E 4π r² = q_{int} / ε₀
E = 1 / 4πε₀ q_{int} / r²
k = 1 /4π ε₀
E = k q_{int} / r² (1)
let's analyze the charge inside the gaussian sphere,
let's use the concept of density for electrons, since they indicate that the charge is evenly distributed
ρ = Q / V
where the volume of the sphere is
V = 4/3 πr³
Qe = ρ V
Qe = ρ 4 / 3π r³
the density of the electrons is
ρ = Ze 3 / (4π R³)
where R is the atomic radius
we substitute
Qe = Ze r³/ R³
for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.
The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)
q_{int} = q_proton + Q_electron
q_{int} = + Ze - Qe
q_{int} = + Ze - Ze r³ / R³
q_{int} = Ze (1- r³ / R³)
we substitute in equation 1
E = k Ze (1- r³ / R³) 1/r²
b) on the surface of the atom r = R
therefore the electric field is zero
E = 0
c) Calculate the electric field for the Uranium for
r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m
E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³ 1/ (5 10⁻¹¹)²
E = -6.62 10¹⁰ N / C
Hello!
Begin by finding the acceleration of the puck. Use the kinematic equation:
The initial velocity is 0 m/s since the object was at rest, so we can rewrite the equation as:
Plug in the given distance and time:
Find the net force using the formula F = m · a (Newton's Second Law)
F = 0.18 · 0.1
F = 0.018 N