I suspect that the pressure of this change is constant therefore
The equation is used from the combined gas law. (When pressure is constant both P's will cancel out P/P = 1)
V/T = V/T
Initial Change
Initially we have 2L at 20 degress what temperature will be at 1L.
2/20 = 1/T
0.1 = 1/T
0.1T = 1
T = 1/0.1
T = 10 degress celsius.
Hope this helps if you won't be able to understand what is the combined gas law just tell me :).
It can only do that when one of the components of the mixture is a magnetic
material.
When you have that situation, you pass the magnet over the mixture ... shaking
the mixture if it's a dry mixture of powders or pieces ... and the magnetic part of
the mixture moves toward the magnet, while the nonmagnetic parts of the mixture
couldn't care less about the magnet and they just stay where they are.
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.
Answer:
D
Explanation:
When a dilution is made, a volume of the stock solution is collected and then is mixed to the solvent. The total amount of the solute (number of moles or mass), must be equal in the volume of the sample and at the final volume, because of the Lavoiser's law (the matter can't be created nor destructed).
The mass or the number of moles is the concentration (C) multiplied by the volume (V), so, if 1 is the sample of the stock solution, and 2 the diluted solution:
C1*V1 = C2*V2
The final volume of the solution is 10 mL. So, let's identify the volume needed for each stock solution.
Acetaminophen
C1 = 500 ug/L
C2 = 50 ug/L
500*V1 = 50*10
V1 = 1 mL
Dextromethorphan
C1 = 1 mM = 1000 uM
C2 = 20 uM
1000*V1 = 20*10
V1 = 0.2 mL
So, the volume of water needed is the total less the volume of the stocks solutions less the volume of the embryo water:
V = 10 - 1 - 1 - 0.2 = 7.8 mL
Thus, to to the solution, it's necessary to add at 1 mL of the embryo water 1 ml of 500 ug/I acetaminophen and 0.2 ml of 1 mM dextromethorphan, and 7.8 ml of water.