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Paladinen [302]
3 years ago
9

Because liquids and gases flow, they are called

Chemistry
1 answer:
kolbaska11 [484]3 years ago
3 0

Answer:

diffusion

Explanation:

You might be interested in
What is the empirical formula of the compound that is 25.3% magnesium and 74.7% chlorine
Margarita [4]

Answer: MgCl_2

Explanation:

25.3% Mg

74.7% Cl

First step: change % to g

25.3g Mg

74.7g Cl

Second step: calculate g/mol of each compound. You can do this by using the atomic mass.

25.3gMg(\frac{1mol}{24.30g})=1.04mol

74.7gCl(\frac{1mol}{35.45g} )=2.11mol

Third step: determine the lowest number and divide everything by it. Of the result, extract whole number only.

Mg=\frac{1.04}{1.04} =1

Cl=\frac{2.11}{1.04}=2

Fourth step: Write each compound with their respective number below.

This empirical formula should be: MgCl_2

4 0
4 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
  • Specific heat of water = 1.035 \frac{cal}{gC}

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

#SPJ1

7 0
2 years ago
Which statement is true of the particles that make up a substance?
victus00 [196]

Particles in a gas have more energy than particles in a liquid. Because in gaseous state particles are free to move around due to which kinetic energy of molecule or gas increases and hence overall energy increases

8 0
3 years ago
7.) A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To
marysya [2.9K]

Answer:

T₂ = 721 k

Explanation:

Given data:

Initial volume = 285 mL

Initial pressure = 1.88 atm

Initial temperature = 355 K

Final temperature = ?

Final volume = 435 mL

Final pressure = 2.50 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂  

T₂  =  P₂V₂ T₁  / P₁V₁

T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL  

T₂ = 386062.5 atm. mL. K /535.8 atm. mL

T₂ = 721 k

5 0
4 years ago
What is the oxidation number of iodine in KL04?<br> (1)- +1<br> (2)- -1<br> (3)- +7<br> (4)- -7
Nostrana [21]
3) +7

I got this stupid question wrong so that you people don't have to. I simply hate chemistry and I wish it didn't exist. Kind of like this website that makes me explain the answer. 

Trust me, the answer is correct
4 0
3 years ago
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