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lisov135 [29]
3 years ago
9

I don't get it Is a Flubber solid, liquid or gas?

Chemistry
2 answers:
Marianna [84]3 years ago
7 0
Flubber would be considered an amorphous solid.<span />
umka2103 [35]3 years ago
5 0
Flubber in this case would be a.... solid, though it might seem he could be a liquid/gas.... Hope it helps ~>~
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Define not matter and give examples​
VLD [36.1K]

Answer:

Everything that has mass and takes up space is matter. Every day, you encounter phenomena that either don't have mass or don't take up space. They are non-matter. Basically, any type of energy or any abstract concept is an example of something that is not matter.

An apple.

A person.

A table.

Air.

Water.

A computer.

Paper.

Iron.

hope this helped you

3 0
3 years ago
A sample of gas has a volume of 215 cm3 at 23.5 degrees Celsius and 84.6kPa what volume will the gas occupy at stp
miss Akunina [59]

The volume of the gas that occupy at STP is 165. 28 cm^3

calculation

by use of combined gas law that is P1V1/T1=P2V2/T2, where

P1=84.6 kpa

T1=23.5 +273=296.5 K

V1=215 cm^3

At STP T= 273 K and P= 101.325 Kpa

therefore p2 = 101.325 Kpa and T2 = 272 K V2=?

by making V2 the subject of the formula V2 =T2P1V1/P2T1

V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3

5 0
3 years ago
Read 2 more answers
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
A disturbance in matter that carries energy from one place to another is called
siniylev [52]
The answer you're looking for is: a wave.
6 0
3 years ago
The brackets are indicating a(n) _____ bond. the brackets are indicating a(n) _____ bond. hydrogen polar covalent single (nonpol
Talja [164]
The brackets are indicating a(n) __Hydrogen___ bond.
1. Hydrogen
3 0
3 years ago
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