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GrogVix [38]
3 years ago
9

The gases listed are at standard temperature and pressure: · 1 mol of CO2 gas · 1 mol of N2 gas · 2 mol of O2 gas Which statemen

t is correct? The volume of 1 mol of CO2 is greater than that of 2 mol of O2 . The volume of 1 mol of CO2 is greater than that of 1 mol of N2 . The volumes of 2 mol of O2 and 1 mol of N2 gases are the same. **The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same.**
Chemistry
2 answers:
andrezito [222]3 years ago
8 0
The right answer for the question that is being asked and shown above is that: "The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same." The <span>statement that is correct as far as the </span>gases listed are at standard temperature and pressure is concerned is that t<span>he volumes of 1 mol of CO2 and 1 mol of N2 gases are the same</span>
Elena L [17]3 years ago
6 0

Answer : The volume of 1 mole of CO_{2} and N_{2} gases are the same.

Solution: Given,

Moles of CO_{2} = 1 mole

Moles of N_{2} = 1 mole

Moles of O_{2} = 2 mole

Formula used for ideal gas is :

P V = n R T

According to the question, the gases are at standard temperature and pressure. So, the volume of gases only depends on the number of moles. This means that the higher the number of moles, higher will be the volume of gas.

The moles of O_{2} are more than the moles of CO_{2} and N_{2}. So, the volume of O_{2} will be more.

And the moles of CO_{2} and N_{2} are equal. Therefore, their volumes are also equal.

Therefore, the best option is the volume of 1 mole of CO_{2} and N_{2} gases are the same.

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STALIN [3.7K]
Yes, K+ is<span> a </span>potassium<span> ion, and Mg</span>2+ is<span> a magnesium ion. But However, when non-metallic elements gain the </span>electrons<span> to form anions, Yes the end of their name </span>is<span> changed to “-ide.” and yes the example, a fluorine </span>atom<span> gains </span>one electron<span> to </span>become<span> a yes fluoride ion (F</span>-<span>) sooo yeessyes</span>
5 0
3 years ago
The "air bags" that are currently installed in automobiles to prevent injuries in the event of a crash are equipped with sodium
stellarik [79]

Answer:

0.0177 L of nitrogen will be produced

Explanation:

The decomposition reaction of sodium azide will be:

2NaN_{3}(s)--->2Na(s)+3N_{2}(g)

As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas

The molecular weight of sodium azide = 65 g/mol

The mass of sodium azide used = 100 g

The moles of sodium azide used = \frac{mass}{molarmass}=\frac{100}{65}=1.54mol

so 1.54 moles of sodium azide will give = \frac{3X1.54}{2}=2.31mol

the volume will be calculated using ideal gas equation

PV=nRT

Where

P = Pressure = 1.00 atm

V = ?

n = moles = 2.31 mol

R = 0.0821 L atm / mol K

T = 25 °C = 298.15 K

Volume = \frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L

3 0
3 years ago
Which of these do not obey the octet rule? select all that apply. select all that apply. clo clo− clo2− clo3− clo4−?
ioda

The correct answer is ClO, ClO3-, ClO- and ClO4-

Kossel and Lewis in 1916 developed an important theory of chemical combination between atoms known as electronic theory of chemical bonding. According to this, atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electron in order to have an octet( 8 electron) in their shells. This is known as octet rule.

In ClO2-, oxygen contains 8 electrons in its valence shell and oxygen will share one electron with chlorine to complete the octet of Cl. In other four, we can clearly see that there are more or less than 8 electrons in the outer shell of oxygen so we can clearly say that ClO, ClO3-, ClO- and ClO4-  are disobeying the octet rule.

3 0
3 years ago
Read 2 more answers
Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
The coordination number of cation and anion in fluorite CaF2 and antifluorite Na2O are respectively​
sp2606 [1]

<u>8:4 and 4:8</u> both have CCP structure............................... ᐛ

6 0
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