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oee [108]
3 years ago
8

A. Sustancia A – Determine el calor específico de una muestra de 1.55 g de acero inoxidable la misma al aumentar la temperatura

a 178°C y que absorbió 141 J.
b. Sustancia B – Determine el calor específico para una muestra de 3.4 g de aceite de oliva a 23°C se le añadió 435 J de calor y aumento su temperatura a 85°C.

II. Utilizando el calor específico de una sustancia (tabla que se encuentra en la presentación) determine la cantidad de calor liberado cuando una muestra de 560 g de agua disminuye la temperatura de 67°C a 22°C.
Chemistry
1 answer:
Ivanshal [37]3 years ago
5 0
Idk tbh nejahxhsjsjns
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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
6 0
2 years ago
The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e
TEA [102]

<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

The work done, W

<u>Explanation:</u>

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

W = -PΔV

Therefore:

ΔH = ΔU - W

W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ


6 0
3 years ago
PLEASE HELP ME NOW I NEED HELP What will happen if a single crystal is introduced into a super-saturated solution? a. The crysta
Klio2033 [76]

Answer:

c

It excess out the crystal

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3 years ago
Which of the following energy does NOT travel as a wave?
frosja888 [35]

Answer:

Wind

Explanation:

It's impossible for them to travel in a wave.

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How many levels will argon have?
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They have about 4 or 6
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