The balanced chemical equation for the reaction is:
2AgNO3 + CaCl2 --> 2AgCl + Ca(NO3)2
Moles of CaCl2 reacted = conc * vol = 0.100 * 3.00 = 0.300 moles
Reaction ratio between CaCl2 and AgNO3 is 1:2. Hence moles of AgNO3 = 0.300 * 2 = 0.600 moles
Volume of AgNO3 = moles / conc = 0.600 / 0.100 = 6.00L
Of course the easy shortcut is that because both solutions are the same concentration you could just multiply 3.00 * 2, however this explanation should reason it out more clearly. Hope I helped you :)
<span> C.The results of the Michelson-Morley experiment did not fit the theory of the luminiferous ether, so the theory had to be rejected. </span>
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.
PH scale is used to determine how acidic or basic a solution is.
we have been given the hydrogen ion concentration. Using this we can calculate pH,
pH = - log[H⁺]
pH = - log (1 x 10⁻¹ M)
pH = 1
using pH can calculate pOH
pH + pOH = 14
pOH = 14 - 1
pOH = 13
using pOH we can calculate the hydroxide ion concentration
pOH = - log [OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 10⁻¹³ M
hydroxide ion concentration is 10⁻¹³ M
