Answer: im not sure im only in 8th grade but im pretty sure Erosion i started learning about this last year i really cant explain it... im still hainving trouble with it.
Explanation:
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
H3PO4 has molecular weight of approximately 98 grams per
mole. 4.50 M is equal to 4.50 mole per 1000 mL solution of H3PO4. 255 mL times
4.50 mol /1000 mL times 98 g/mol is equal to 112.455 grams. Note that I
automatically equate 1 Liter to 1000 mL since the given volume is in mL for
easier computation.
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.
where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.
It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.
The oxidation state of Pb in 
is 2. So, moles deposited by Pb is as follows.
It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.
Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
