Answer:
1.06 mL of toluene will be needed.
Explanation:
Equi-molar mixture means equal moles of all the components.
as given the volume of ethyl acetate = 1mL
Density of ethyl acetate = 0.898 g/mL
The relation between density, mass and volume is :
mass=volumeXdensity
mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams
the moles are related to mass as:
For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol
moles of ethyl acetate will be:
So we need 0.01 moles of toluene also
For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene
mass required = 0.01 X 92=0.92grams
for 0.92 grams of toluene volume required will be:
Instrumental methods of analysis rely on machines.The visualization of single molecules, single biological cells, biological tissues and nanomaterials is very important and attractive approach in analytical science.
There are several different types of instrumental analysis. Some are suitable for detecting and identifying elements, while others are better suited to compounds. In general, instrumental methods of analysis are:
-Fast
-Accurate (they reliably identify elements and compounds)
-Sensitive (they can detect very small amounts of a substance in a small amount of sample)
Answer:
127.0665 amu
Explanation:
Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.
We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g
The percentage abundances is as follows:
Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%
Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%
Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.
That is :
[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M
1) chromium(III) nitrate is acidic, because it is the salt of weak base (chromium(III) hydroxide Cr(OH)₃) and strong acid (nitric acid HNO₃).
2) sodium hydrosulfide is basic, because it is the salt of strong base (sodium hydroxide MaOH) and weak acid (hydrogen sulfide H₂S).
3) zinc acetate is little basic, because zinc hydroxide (Zn(OH)₂) is stronger base than acetic acid (CH₃COOH).
