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katrin2010 [14]
3 years ago
7

The compound XCl is classified as Ionic if x represents the element?

Chemistry
2 answers:
puteri [66]3 years ago
6 0
Well since chlorine upon receiving a valence electron from the X element, we would know that the element would be a metal, based on the charge of Cl, after it receives a valence electron is -1, the metal would have to give 1 valence electron as well, and thus the possible elements can be any metal within the group 1 of the periodic table of elements. Lithium, Potassium etc.
Studentka2010 [4]3 years ago
6 0

Answer:

X belongs to the alkali metals which can be lithium, sodium, potassium, rubidium and caesium.

Explanation:

Ionic bond: This type of bond is formed when there is a complete transfer of electrons from one element to another element. In this bonding one element is always a metal and another is a non-metal.

Chlorine is the element of group 17 and third period. The atomic number of chlorine is 17 and the symbol of the element is Cl.

The electronic configuration of the element chlorine is:-

1s^22s^22p^63s^23p^5

Thus, chlorine need one electron to become stable and form ionic bond.

Also, the ionic compound has the formula XCl which means that the valency of X is +1.

<u>Thus X belongs to the alkali metals which can be lithium, sodium, potassium, rubidium and caesium.</u>

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<h3>Pacing Study Sessions</h3>

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The correct answer is:

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5 0
2 years ago
What does a gamma ray primarily consist of?
damaskus [11]
Answer is D

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6 0
3 years ago
An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will
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Answer:

It will be halve of T

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6 0
3 years ago
Which direction does air circulate into low pressure zones in the northern and southern hemisphere
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5 0
3 years ago
Read 2 more answers
Calculate δg o for each reaction using δg of values:(a) h2(g) + i2(s) → 2hi(g) kj (b) mno2(s) + 2co(g) → mn(s) + 2co2(g) kj (c)
steposvetlana [31]
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol

Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol

Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
      = (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
      = 176.15 kJ - 84.78 kJ = 91.38 kJ 
 





 
6 0
3 years ago
Read 2 more answers
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