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77julia77 [94]
3 years ago
6

For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou

ld NOT be the change in enthalpy (ΔH) and the activation energy (Ea), respectively, for the catalyzed reaction? Select all that apply.
Physics
1 answer:
Ronch [10]3 years ago
5 0

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

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2 years ago
A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c
Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

Q=mC_s \Delta T

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

\Delta T=40.0 C - 80.0 C=-40.0 C is the change in temperature of the wire

Substituting into the equation, we find

Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J

And the sign is negative because the heat is released by the wire.

6 0
3 years ago
What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?
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A

Explanation:

The earth is spherical. So it's middle part is bulgjng outside. So more sunlight will be incident on the latitude near the equator. This will heat the air and it will rise up. This will cause high pressure difference and polar disturbances.

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3 years ago
What types of collisions can result from making unsafe passes?
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3 years ago
A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

(a) a=2m/sec^2

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2

(a) We know that force is given by

F = ma

So force will be F=290\times 2=580N

(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

5 0
3 years ago
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