Do the following ! I'LL MARK THE BRAINLEST!<br> ITS SCIENCE!

Chapter 5. You must show all your work. Solve the following problems. (a) (8 points) When a cold drink is taken from a refrigera

GuDViN [60]

Answer:

see explanation below

Explanation:

To do this, we need to use the Newton's law of cooling which is:

dT/dt = (k - Ts)

Where:

Ts: temperature of surroundings (In this case, 20 °C)

k: constant

t: time

Now, after we do the integrals of this, the general expression would be:

T(t) = Ce^kt + Ts (1)

Now, with the first data, we need to calculate the value of C. This value will be the same after time has passed. You can see this as the concentration of the drink. As it's not experimenting any reaction, it's concentration remain the same, and only the temperature will change.

Now, to get the value of C, we can begin with the fact that at time = 0, temperature was 5°C so:

T(0) = Ce^k*0 + Ts

Replacing:

5 = Ce^1 + 20

5 - 20 = C*1

C = -15

Now that we have this, we can solve the first part of the problem

(i):

First, we need to get the value of k, we know the final temperature at t = 25, so we can solve for k, which is constant too, and then, calculate the temperature for t = 50 min

solving for k, with T = 10 °C, C = -15, Ts = 20 °C and t = 25 min:

10 = -15e^25k + 20

10 - 20 = -15e^25k

-10/-15 = e^25k

ln(-10/-15) = 25k

k = -0.405465/25

Now that we have k, let's calculate T after t = 50

T = -15e^(-0.0162)*50 + 20

T = -6.67 + 20

T(50) = 13.33 °C

(ii)

For this part, we only need to solve for t:

18 = -15e^(-0.0162)t + 20

18 - 20 / -15 = e^-0.0162t

0.1333 = e^-0.0162t

ln(0.1333) = -0.0162t

t = -2.0145/-0.0162

t = 124.37 min

8 0

3 years ago

Calculate the equilibrium constant for the reaction: 2 Cr + 3 Pb2+ ----> 3 Pb + 2 Cr3+ at 25oC. Eocell = 0.61 V

sattari [20]

Answer:

The value is $K = 8*10^{61}$

Explanation:

From the question we are told that

The equation is $2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}$

The temperature is $T = 25^oC = 298 K [room \ temperature ]$

The emf at standard condition is $E^o_{cell} = 0.61 \ V$

Generally at the cathode

$3Pb^{2+}(aq) + 6 e- --> 3Pb(s)$

At the anode

$2Cr^{3+} + 6e^- \to 2Cr$

Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as

$G = n* F * E^o_{cell}$

Here n is the no of electron with value n = 6

F is the Faraday's constant with value 96487 J/V

=> $G = 6 * 96487 * 0.61$

=> $G = 3.5 *10^{5} \ J$

This Gibbs free energy can also be represented mathematically as

$G = RTlogK$

Here R is the cell constant with value 8.314J/K

K is the equilibrium constant

From above

=> $K = antilog^{\frac{G}{ RT} }$

Generally antilog = 2.718

=> $K = 2.718^{\frac{3.5 *10^5}{ 8.314* 298} }$

=> $K = 8*10^{61}$

6 0

3 years ago

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem

Ludmilka [50]

Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L

⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K

⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

5.10 L / 471 K = 2.50 L / T2

T2 = 2.50 L / (5.10 L / 471 K)

T2 = 230.9 K = -42.1

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

8 0

3 years ago

The temperature at which a substance in the liquid phase transforms to the gaseous phase refers to the substance's _______. A. s

murzikaleks [220]

When a substance goes from being a liquid to a gas it evaporates, or boils away. Think of boiled eggs.

3 0

3 years ago

Read 2 more answers

If the temperature of 15 grams of water changes from 21C to 24C, how many joules of heat were involved? Show work

goblinko [34]

Answer:

189 Joules

Explanation:

Applying,

Q = cm(t₂-t₁)............. equation 1

Where Q = Heat, c = specific heat capacity of water, m = mass of water, t₁ = Initial Temperature, t₂ = Final temperature.

From the question,

Given: m = 15 grams = 0.015 kg, t₁ = 21 °C, t₂ = 24 °C

Constant: c = 4200J/kg.°C

Substitute these values into equation 1

Q = 0.015×4200×(24-21)

Q = 0.015×4200×3

Q = 189 Joules

6 0

3 years ago

Do the following ! I'LL MARK THE BRAINLEST! ITS SCIENCE!