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suter [353]
3 years ago
5

State one use for argon.

Chemistry
2 answers:
Alex17521 [72]3 years ago
8 0

Argon is perticularly important for the metal industry, being used as an inert gas shield in arc

Colt1911 [192]3 years ago
5 0
Non-reactive blanket in the manufacture of titanium and other reactive elements and as a protective atmosphere for growing silicon and germanium crystals.
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Phosphorus atomic radius is smaller than magnesium atomic radius <br> True or false
stealth61 [152]

Answer:

True

Explanation:

Atomic radius can be defined as a measure of the size (distance) of the atom of a chemical element such as hydrogen, oxygen, carbon, nitrogen etc, typically from the nucleus to the valence electrons. The atomic radius of a chemical element decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon). Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).

<em>Hence, the atomic radius of phosphorus is smaller than the atomic radius of magnesium. Basically, the atomic radius of phosphorus is 98 pm while the atomic radius of magnesium is 145 pm.</em>

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3 years ago
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What type of bond does hydrogen and carbon form
laila [671]
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7 0
3 years ago
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As you move from left to right across the periodic table elements ?
Sidana [21]

Explanation:

Quite a number of properties varies across a period. Some remains constant whereas others decreases.

As one moves from left to right;

  • The energy level remains the same.
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6 0
3 years ago
A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
bija089 [108]

Answer:

E = 0.062 V

Explanation:

(a) See the attached file for the answer

(b)

Calculating the voltage (E) using the formula;

E = - (2.303RT/nf)log Cathode/Anode

Where,

R = 8.314 J/K/mol

T = 35°C = 308 K

F- Faraday's constant = 96500 C/mol,

n = number of moles of electron = 2

Substituting, we have

E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)

   = -0.031 * -2

  = 0.062V

Therefore, the voltmeter will show a voltage of 0.062 V

5 0
3 years ago
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