Answer: Saturated
A solution that contains the maximum amount of solute that is soluble at a given temperature is said to be saturated.
Explanation:
A Saturated solution is one that contains as much (i.e maximum) solute as it can dissolve at that temperature in the presence of undissolved solute particles.
For instance: if a given volume of water can only dissolve a certain amount of salt in it at room temperature, then, more salt added will not dissolve.
Thus, making the solution saturated.
Ionization Trend: First ionization energy will increase left to right across a period and increase bottom to top of a family (column).
A) Sr, Be, Mg are all in column 2 of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Be, Mg, Sr
B) Bi, Cs, Ba are all in the same row of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Bi, Ba, Cs
C) Same rule as above. Order is: Na, Al, S
Answer:regulating body temperature
Explanation:
The question is incomplete. The complete question is:
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:
ΔV = x1x2(45x1 + 25x2)
Where ΔV is in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and 2VE in a mixture containing 40 mole percent of specie 1.
Answer:
1VE = 117.92 cm³.mol⁻¹, 2VE = 97.92 cm³.mol⁻¹
Explanation:
In the equation given, x represents the molar fraction of each substance, thus x1 = 0.4 and x2 = 0.6. Because of the mixture, the molar partial volume of each substance will change by a same amount, which will be:
ΔV = 0.4*0.6(45*0.4+ 25*0.6)
ΔV = 7.92 cm³.mol⁻¹
1VE - V1 = 7.92
1VE = 7.92 + 110
1VE = 117.92 cm³.mol⁻¹
2VE - V2 = 7.92
2VE = 7.92 + 90
2VE = 97.92 cm³.mol⁻¹
Answer:
0.124 M
Explanation:
The reaction obeys second-order kinetics:
![r = k[BrO^-]^2](https://tex.z-dn.net/?f=r%20%3D%20k%5BBrO%5E-%5D%5E2)
According to the integrated second-order rate law, we may rewrite the rate law in terms of:
![\dfrac{1}{[BrO^-]_t} = kt + \dfrac{1}{[BrO^-]_o}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_t%7D%20%3D%20kt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D)
Here:
is a rate constant,
![[BrO^-]_t](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t)
is the molarity of the reactant at time t,
![[BrO^-]_o](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_o)
is the initial molarity of the reactant.
Converting the time into seconds (since the rate constant has seconds in its units), we obtain:
Rearranging the integrated equation for the amount at time t:
![[BrO^-]_t = \dfrac{1}{kt + \dfrac{1}{[BrO^-]_o}}](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7Bkt%20%2B%20%5Cdfrac%7B1%7D%7B%5BBrO%5E-%5D_o%7D%7D)
We may now substitute the data:
![[BrO^-]_t = \dfrac{1}{0.056 M^{-1}s^{-1}\cdot 60.0 s + \dfrac{1}{0.212 M}} = 0.124 M](https://tex.z-dn.net/?f=%5BBrO%5E-%5D_t%20%3D%20%5Cdfrac%7B1%7D%7B0.056%20M%5E%7B-1%7Ds%5E%7B-1%7D%5Ccdot%2060.0%20s%20%2B%20%5Cdfrac%7B1%7D%7B0.212%20M%7D%7D%20%3D%200.124%20M)
