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3 years ago

How to turn any drink into a gummy

Chemistry

1 answer:

Alika [10]3 years ago

5 0

Add gelatin and let it boil, then put it in the freezer for 30 minuets

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What is the charge of the nucleus in an Adam of oxygen -17

Ivenika [448]

Charge of nucleus is always positive whether it is element or isotope.

5 0

3 years ago

A gas sample is held at constant temperature. What happens to the pressure of the sample if the volume is doubled and the number

Iteru [2.4K]

Answer:

If the volume is doubled and the number of molecules is doubled, pressure is unchanged

Explanation:

Step 1: Data given

Temperature = constant

Volume will be doubled

Number of molecules will be doubles

Step 2:

p*V = n*R*T

⇒ gas constant and temperature are constant

Initial pressure = n*R*T / V

Initial pressure = 2*R*T/2

Initial pressure = RT

Final pressure = 4*RT / 4

Final pressure = R*T

If the volume is doubled and the number of molecules is doubled, pressure is unchanged

3 0

3 years ago

1. Stirring increases the rate of solution

Blababa [14]

Answer:

1) The correct answer is A. Collision

2) A hot solvent helps a solid dissolve faster because an increase in kinetic energy that also increases the rate of collisions

Explanation:

When a solute is added into a solvent and stirred, the solute particles get distributed to all parts of the solvent as a result of stirring.

More collisions occur between the solute and the solvent due to stirring. This increases the rate of dissolving.

When a solvent is heated, then the kinetic energy would increase and the atoms will collide with a much greater force. As a result, ore solute will be able to dissolve in the solvent.

3 0

3 years ago

One sphere has a radions of 181 cm, another has a radius of 5.01 cm. What is the difference in volume (in cubic centimeters) bet

PSYCHO15rus [73]

Answer:

$2.4*10^{7}cm^{3}$

Explanation:

First you should calculate the volume of a big sphere,so:

$V_{big}=\frac{4}{3}\pi r^{3}$

$V_{big}=\frac{4}{3}\pi (181cm)^{3}$

$V_{big}=2.4*10^{7}cm^{3}$

Then you calculate the volume of a small spehre, so:

$V_{small}=\frac{4}{3}\pi r^{3}$

$V_{small}=\frac{4}{3}\pi (5.01cm)^{3}$

$V_{small}=5.3*10^{2}cm^{3}$

Finally you subtract the two quantities:

$V_{big}-V_{small}=2.4*10^{7}cm^{3}-5.3*10^{2}cm^{3}$

$V_{big}-V_{small}=2.4*10^{7}cm^{3}$

5 0

3 years ago

Without constants you would not know which variable affected the

bonufazy [111]

Two independent variables could change at the same time, and you would not know which variable affected the dependent variable

4 0

3 years ago