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2 years ago

100 g of water are cooled and the enthalpy change is -12,540 J. What is the change in temperature of the water?

Physics

1 answer:

Gnesinka [82]2 years ago

5 0

Answer:

-29.9907

Explanation:

c = Q / (mΔT)

You're looking for ΔT, so you would plug in what you already know and get C= -12540/(100 x ΔT)

then you would solve

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

Ronch [10]

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

Draw Graph for the derivation of three equations of Motion

Over [174]

Answer:

Hey mate I shall not tell you the answer I shall explain it to you after this if still you can't understand then say

Explanation:

Derive v = u + at by Graphical Method. Consider the velocity – time graph of a body shown in the below Figure

Derive s = ut + (1/2) at2 by Graphical Method. Velocity so time graph to derive the equations of motion.

Derive v2 = u2 + 2as by Graphical Method. Velocity–Time graph to derive the equations of motion.

I hope you understand now

enjoy your day

#Captainpower :)❤❤

5 0

2 years ago

Which statement is TRUE?

Salsk061 [2.6K]

Statement B is true.

3 0

3 years ago

Read 2 more answers

The chef at the infamous Fattening Tower of Pizza tosses a spinning disk of uncooked pizza dough into the air. The disk becomes

Bess [88]

Rotational speed would increase...

v = omega . r

which means it's directly proportional to radius...

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3 years ago

A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?

jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

$Total\hspace{3}distance=2*d$

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

$S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h$

5 0

2 years ago