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lara [203]
3 years ago
11

Which processes represent one chemical change in one physical change

Chemistry
2 answers:
prisoha [69]3 years ago
7 0

 decomposition and melting

Alex Ar [27]3 years ago
4 0

Answer:

Decomposition and melting

Explanation:

Decomposition is the process of splitting two or more compounds apart from each other. Usually, the decomposed substance have different chemicals identities hence the reason why it is a chemical process since it causes chemical change. Example is the decomposition of hydrated salts of metals. CuSO4.5H20 is a penta-hydrate salt I.e contains 5 molecules of water. If this compound is heated, two different compounds are formed and changes the chemical identity of the compound. Melting is a process of liquidizing compounds or elements to the point it changes physical state from solid-liquid. Melting does not change the chemical identity of the compound but rather the physical state of the compound.

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A plastic bin is found to hold 3.1x10^24 molecules of water.
motikmotik

Answer:

\boxed {\boxed {\sf 5.1 \ mol \ H_2O}}

Explanation:

To convert from representative particles to moles, Avogadro's Number: 6.02*10²³, which tells us the number of particles (atoms, molecules, etc.) in 1 mole of a substance.

We can use it in a ratio.

\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Multiply by the given number of molecules.

3.1*10^{24} \ molecules \ H_2O*\frac {6.02*10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Flip the ratio so the molecules of water cancel out.

3.1*10^{24} \ molecules \ H_2O*\frac {1 \ mol \ H_2O}{6.02*10^{23} \ molecules \ H_2O}

3.1*10^{24} *\frac {1 \ mol \ H_2O}{6.02*10^{23} }

\frac {3.1*10^{24} \ mol \ H_2O}{6.02*10^{23} }

Divide.

5.14950166113 \ mol \ H_2O

The original number of molecules has 2 significant figures: 3 and 1, so our answer must have the same. For the number we calculated, that is the tenth place. The 4 in the hundredth place tells us to leave the 1.

5.1 \ mol \ H_2O

There are about 5.1 moles of water in 3.1*10²⁴ molecules of water.

5 0
2 years ago
the density of air at ordinary atmospheric pressure and 25 degrees celcius is 1.19 g/L. what is the mass, in kilograms, of the a
riadik2000 [5.3K]
The mass of the air in the room is 50 kg.

V = lwh = 15.5 ft × 12.5 ft × 8 ft = 1550 ft³

1 ft = 12 in × \frac{2.54 cm}{1 in} = 30.48 cm

V = 1550 ft³ ×(\frac{30.48 cm}{1 ft})^{3} = 4.39 × 10⁷ cm ³ = 4.39 × 10⁷mL = 4.39 × 10⁴ L

Mass = 4.39 × 10⁴ L × \frac{1.119 g}{1 L} = 5.22 × 10⁴ g = 52.2 kg

Note: The answer can have only 1 significant figure, because that is all you gave for the height of the room.

To the correct number of significant figures, the mass of the air in the room is 50 kg.
6 0
3 years ago
Which scientist established the law of octaves?
laiz [17]
He was an english scientist named john newlands
6 0
3 years ago
Read 2 more answers
3. If a sound is louder or a light is brighter, what does this mean about the amplitude of
hodyreva [135]

Answer: for 3. I would say it means that the amplitude is greater because if a sound is louder that means the amplitude is larger, same with the light.So that would mean the softer the sound the smaller the amplitude same with the dim of the light so for 4. It would be the amplitude is smaller. Hope this helps! :) ( if you have any questions feel free to ask)

7 0
3 years ago
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
  • The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
  • 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
  • The molar mass of silver is 107.87 g/mol

The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
3 years ago
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