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valina [46]
3 years ago
8

State the number of electrons that must be gained by atoms of each of the following elements to obtain a stable electron configu

ration. a) CI b) Se C) N D) I E) S
Chemistry
1 answer:
Molodets [167]3 years ago
8 0

Answer:

The answer to your question is below

Explanation:

a) CI Chlorine is located in group VIIA, so its must gain one electron to be stable.

b) Se Selenium is located in group VIA, it must gain 2 electrons to be stable.

c) N Nitrogen is located in group VA, it must gain 3 electrons to be stable.

d) I Iodine is located in group VIIA, so it must gain 1 electron to be stable.

e) S sulfur is located in group VIA, so it must gain 2 electrons to be stable.

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A mixture of gases has 0.3000 mol of CO2, 0.2706 mol of SO2 and 0.3500 mol of water vapor. The total pressure of the mixture is
notka56 [123]

Answer:

Partial pressure SO₂ → 0.440 atm

Explanation:

We apply the mole fraction concept to solve this:

Moles of gas / Total moles = Partial pressure of the gas / Total pressure

Total moles = 0.3 moles of CO₂ + 0.2706 moles of SO₂ + 0.35 moles H₂O

Total moles = 0.9206 moles

Mole fraction SO₂ = 0.2706 moles / 0.9206 moles → 0.29

Now, we can know the partial pressure:

0.29 = Partial pressure SO₂ / Total pressure

0.29 = Partial pressure SO₂ / 1.5 atm

0.29 . 1.5atm = Partial pressure SO₂ → 0.440 atm

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How do you know flammability is a chemical property?
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Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

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Therefore, the enthalpy change for C_2H_2 is -227 kJ.

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