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Answer:
(a) ![102.6g/mol](https://tex.z-dn.net/?f=102.6g%2Fmol)
(b) Rubidium
Explanation:
Hello,
This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:
![V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}](https://tex.z-dn.net/?f=V_%7Bbase%7D%3D0.1L%3B%20M_%7Bacid%7D%3D2.5M%2C%20V_%7Bacid%7D%3D0.017L%5C%5CV_%7Bbase%7DM_%7Bbase%7D%3DV_%7Bacid%7DM_%7Bacid%7D)
Solving for the molarity of base we've got:
![M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L](https://tex.z-dn.net/?f=M_%7Bbase%7D%3D%5Cfrac%7BM_%7Bacid%7D%2AV_%7Bacid%7D%7D%7BV_%7Bbase%7D%7D%3D%5Cfrac%7B2.50M%2A0.017L%7D%7B0.1L%7D%20%3D0.425M%3D0.425mol%2FL)
Now, we can compute the moles of the base as:
![n_{base}=0.425mol/L*0.1L=0.0425mol](https://tex.z-dn.net/?f=n_%7Bbase%7D%3D0.425mol%2FL%2A0.1L%3D0.0425mol)
(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:
![\frac{4.36g}{0.0425mol} =102.6g/mol](https://tex.z-dn.net/?f=%5Cfrac%7B4.36g%7D%7B0.0425mol%7D%20%3D102.6g%2Fmol)
(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:
![102.6g/mol-16g/mol-1g/mol=85.6g/mol](https://tex.z-dn.net/?f=102.6g%2Fmol-16g%2Fmol-1g%2Fmol%3D85.6g%2Fmol)
So, that atomic mass dovetails to the Rubidium's atomic mass.
Best regards.
Answer: 0.000625
Explanation:
If you don't know, one Milligram is equivalent to 0.000001 Kilograms. Hence, we'll multiply 625 by 0.000001.
With that being said, 625 times 0.000001 will equal to 0.000625.
Don't forget to add the units.
Milogram unit ⇒ mg
Kilogram unit ⇒ kg
Finally, your grand answer is 0.000625.
<em>Please comment down below for any questions about my answer.</em>
Answer:
A
Explanation:
Am Inference is an educated guess, or a guess using prior knowledge
Answer:
a, d
Explanation:
a. The pressure in the container will be 42.8 mm Hg.
d. Some of the vapor initially present will condense.