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Vlada
3 months ago
5

Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g

iven: 2a + b → a2b δh°
Physics
1 answer:
Softa [3K]3 months ago
3 0

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

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Answer:

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Explanation:

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E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}

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4 0
3 months ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
ValentinkaMS [3465]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C represents the charge's magnitude

and \Delta V = 6.0 kV = 6000 V signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
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Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s

Taking a beam diameter of 2 mm = 2 x 10^-3 m

therefore

the beam's cross-sectional area A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) The intensity I = P/A

where P refers to the laser's power

and A represents the beam's cross-sectional area

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) The total energy delivered E =Pt

where P is the beam's power

and t is the exposure duration

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field can be computed as

E = \sqrt{2I/ce_{0} }

where I signifies the beam's intensity

and E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9} = 1.55 x 10^-8 v/m

6 0
2 months ago
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Answer:

E_t_o_t_a_l=7.603MJ

Explanation:

The overall energy expenditure of the salmon, which corresponds to its swimming upstream effort, W, is linked to its specific mechanical power. Mechanical \ power calculated per unit mass can be derived from the following equation:

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As a result, the total energy utilized during the 22-day journey is 7.603 MJ

6 0
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