Answer:
a salt
Explanation:
compounds are made of molecules but we need to mention the type of compound formed.
Answer:
(a) The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.
Explanation:
Q is the coefficient of the reaction and is calculated the same of the way of the equilibrium constant, but using the concentrations or partial pressures in any moment of the reaction, so, for the reaction given:
Q = (pBrCl)²/(pBr₂*pCl₂)
Q = 2²/(1x1)
Q = 4
As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.
The partial pressures of Br₂ and Cl₂ must decrease, so they will be smaller than 1.00 atm. And the total pressure must not change because of the stoichiometry of the reaction: there are 2 moles of the gas reactants for 2 moles of the gas products.
Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.
I believe it would be A, frogs and spiders, because frogs and spiders both eat grasshoppers. But I cannot see the diagram it talks about, so it's harder to tell.
Answer : The entropy change of reaction for 1.62 moles of 
reacts at standard condition is 217.68 J/K
Explanation :
The given balanced reaction is,
The expression used for entropy change of reaction 
is:
![\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BBr_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28Br_2%29%7D%2Bn_%7BF_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28F_2%29%7D%5D-%5Bn_%7BBrF_3%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28BrF_3%29%7D%5D)
where,
= entropy change of reaction = ?
n = number of moles
= standard entropy of formation
= 245.463 J/mol.K
= 202.78 J/mol.K
= 292.53 J/mol.K
Now put all the given values in this expression, we get:
![\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B1mole%5Ctimes%20%28245.463J%2FK.mole%29%2B3mole%5Ctimes%20%28202.78J%2FK.mole%29%7D%5D-%5B2mole%5Ctimes%20%28292.53J%2FK.mole%29%5D)
Now we have to calculate the entropy change of reaction for 1.62 moles of reacts at standard condition.
From the reaction we conclude that,
As, 2 moles of has entropy change = 268.74 J/K
So, 1.62 moles of has entropy change = 
Therefore, the entropy change of reaction for 1.62 moles of reacts at standard condition is 217.68 J/K
Hey there!
The number of vacancies per unit volume => ( Nv = 2.3*10²⁵ m⁻³ )
Avogrado's number => ( NA = 6.022*10²³ atoms/mol )
Density of material ( p ) in g/m³ :
1 g/cm³ = 1000000 g/m³ so:
7.40 * ( 1000000 ) = 7.40*10⁶ g/m³
Atomic mass = 85.5 g/mol
* Calculate the number of atomic sites per unit volume :
N = NA * p / A
N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5
N = 4.45*10³⁰ / 85.5
N = 5.212*10²⁸ atoms/m³
Therefore:
Calculate the fraction of vacancies :
Fv = Nv / N
Fv = 2.3*10²⁵ / 5.212*10²⁸
FV = 4.441*10⁻⁴
Hope that helps!
