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romanna [79]
3 years ago
11

How do you use nonmetal in a sentence?

Chemistry
1 answer:
skad [1K]3 years ago
7 0
Here Is A Sentence You Could Possibly Use:

"Hydrogen and Nitrogen are two examples of non-metals."

(You can replace the two given elements with whatever two non-metallic elements you like.)

Hope This Helps!

BTW, You May Want A Second Opinion Just In Case! :)
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List three changes of state during which energy is absorbed
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"<span>Changes of state are physical changes. They occur when matter absorbs or loses energy. Processes in which matter changes between liquid and solid states are freezing and </span>melting<span>. Processes in which matter changes between liquid and gaseous states are vaporization, evaporation, and condensation."</span>
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One radioactive isotope of calcium has an atomic mass of 47,how many neutrons are in the calcium -47 nucleus
jekas [21]

Answer: 27 neutrons

Explanation:

Recall that the number of protons in the nucleus of an atom = Atomic number.

Hence, Calcium with mass number 47 and atomic number 20 will have 20 protons

Therefore, since Mass number = number of protons + neutrons

47 = 20 + neutrons

Neutrons = 47 - 20 = 27

Thus, there are 27 neutrons in radioactive calcium nucleus.

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All of the following will be damaging to the environment except
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Adding fertilizer to stimulate the aquatic plant growth, and putting research and money into renewable energy
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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

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What type of energy results from the
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