Please hurry!!! True or false? In a chemical reaction, energy may be gained or lost but the atoms stay the same

An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial

andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state ) P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm

P₂= 2.25 atm

8 0

2 years ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Mrrafil [7]

Answer :

The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.

Percent Ionic Character :

It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom

Given : Molecule is TiAl₃

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 ( From image shared )

Plug the value in above formula :

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

Value of e⁻¹ = 0.90

Percent ionic character = 1 - 0.90 * 100

Percent Ionic character = 10 %

Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .

8 0

2 years ago

An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or

galben [10]

The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s

4 0

2 years ago

Read 2 more answers

N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t

son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.

Explanation : When nitrogen ( $N_{2}$ )and hydrogen ( $H_{2}$ )are mixed, the mole ratio becomes 1 : 1.5,