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Norma-Jean [14]
3 years ago
6

How does the amount of space between particles differ in the ballon and helium cylinder

Chemistry
1 answer:
Ivanshal [37]3 years ago
8 0
In a balloon there is a low pressure within the balloon. This means that the particles are more spaced out. Wheras in the cylinder they are under high pressure, meaning that the particles are much closer together!
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Identify each process numbered 1-6
Vedmedyk [2.9K]

Answer:

1 = Melting

2 = Freezing

3 = Sublimation

4 = Deposition

5 = Condensation

6 = Boiling

Explanation:

7 0
3 years ago
Substance A is a nonpolar liquid and has only dispersion forces amongits constituent particles. Substance B is also a nonpolar l
alekssr [168]

Answer:

(a) when both the liquid are mixed, they become completely miscible as the intermolecular forces between them are equal to the intra molecular forces

(b) since the magnitude of intermolecular force = intra molecular forces, Delta H solution=0

(c) Delta H solute = positive

Delta H solvent = positive

Delta H mix = negative

Explanation:

3 0
3 years ago
What is the difference between an element and a particle?
soldier1979 [14.2K]
Elements are made up of particles.
5 0
3 years ago
HELP.
Kamila [148]

Answer:

The mass of tin is 164 grams

Explanation:

Step 1: Data given

Specific heat heat of tin = 0.222 J/g°C

The initial temeprature of tin = 80.0 °C

Mass of water = 100.0 grams

The specific heat of water = 4.184 J/g°C

Initial temperature = 30.0 °C

The final temperature = 34.0 °C

Step 2: Calculate the mass of tin

Heat lost = heat gained

Qlost = -Qgained

Qtin = -Qwater

Q = m*c*ΔT

m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)

⇒with m(tin) = the mass of tin = TO BE DETERMINED

⇒with c(tin) = the specific heat of tin = 0.222J/g°C

⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C

⇒with m(water) = the mass of water = 100.0 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C

m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C

m(tin) =  163.9 grams ≈ 164 grams

The mass of tin is 164 grams

4 0
3 years ago
Consider the reaction. CaCl2(aq)+K2CO3(aq)⟶CaCO3+2KCl. Identify the precipitate, or lack thereof, for the reaction. (A) KCl (B)
Natali5045456 [20]

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

Precipitate is defined as insoluble solid substance that emerges when two different aqueous solutions are mixed together. It usually settles down at the bottom of the solution after sometime.

For the given chemical equation:

CaCl_2(aq.)+K_2CO_3(aq.)\rightarrow CaCO_3(s)+2KCl(aq.)

The products formed in the reaction are calcium carbonate and potassium chloride. Out of the two products, one of them is insoluble which is calcium carbonate. Thus, it is considered as a precipitate.

Hence, the correct answer is Option B.

4 0
4 years ago
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