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sleet_krkn [62]
3 years ago
13

Does an object that feels cold to the touch contain thermal energy? explain your answer.

Chemistry
1 answer:
Len [333]3 years ago
3 0
YES - explained (Temperature is an average of the kinetic energy isn't it? Thermal energy is the total. So you could have a really big "cold" object and it would have more energy then say a small "hot" object. But the average energy of the hot object would be greater than the average of the cold object. "You only need to know the amount of thermal energy a body contains to calculate its temperature" -You need to know something about the size of the objects and the particles that made it up? "The object contains less thermal energy than your hand" -It could be a large cold object, so it could have a greater amount of energy but over a larger area, so it's just the temperature that is lower. I've only studied thermodynamics a little, but I think that's sorta right...)
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Answer:

The particles in a solid are tightly packed and locked in place. ... The particles in a liquid are close together (touching) but they are able to move/slide/flow past each other. The particles in a gas are fast moving and are able to spread apart from each other.

Explanation:

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Why is thre no gas bubble whene there is citric acid?answer all the questions in detail.​
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Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−
kolbaska11 [484]

Answer:

r = 3.61x10^{-6} M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.[S2O2^{-8} ]^{x} x [I^{-} ]^{y}

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}

\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y

0.67 = 0.67^x

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}

\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y

0.33 = 0.5x 0.67^y

0.67 = 0.67^y

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

2.6x10^{-6} = kx0.018x0.036kx6.48x10^{-4} = 2.6x10^{-6}

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = 4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}

r = 3.61x10^{-6} M/s

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