All categories

3 years ago

A beverage manufacturer wants to increase the solubility of carbon dioxide (CO2) in its carbonated drinks.

Physics

1 answer:

Brut [27]3 years ago

6 0

D. Decreasing its temperature

Explanation:

Decreasing the temperature of the carbon dioxide gas to be dissolved in the carbonated drink will most likely increase the solubility of the gas in the drink.

Temperature has considerable effects on the solubility of gases in liquids.

Dissolution involves the surrounding of ions by water molecules, in this case, the carbon dioxide gas is to be surrounded by the liquid beverage medium.
Increasing pressure increases the rate at which gases are soluble. At high pressure, the gases are brought more in contact with the liquid medium.
Decreasing temperature aids gas solubility.
If the temperature of gases are increased, they will not want to stay in solution as they gain a high amount of kinetic energy.
Therefore, it will increase their randomness and the urge to leave the solution.
Decrease in temperature and increase in pressure makes gas solubility to be fast.

Learn more:

Rate of chemical reactions brainly.com/question/6281756

#learnwithBrainly

You might be interested in

HEY CAN YALL PLS HELP ME IN DIS!!!!!!!!!!!!!

Serga [27]

Answer:

Hey!!

Your answer is: 0.72

Explanation:

if 760=1 then...

550=x

=550÷760= 0.72 in two s.f

6 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$