I think this is a units issue, i keep getting the wrong answer.
the formula should be rate1/rate2= (molar mas 2/ molar mass1)^1/2
but every time i do it, i get it wrong, and when i do rate1/rate2= (molar mas 1/ molar mass2)^1/2, i get the right answer.
It takes 140 SECONDS for a sample of neon to effuse through a porus plug, how long does it take for the same volume of butane, C4H10 to effuse under the same conditions?
do i have to change seconds to anything? What unit does the law say rate is in?
Please help!
If you get it "right" the second way, the machine determining right is in error.
rateH2/rateNe=sqrt (20/2) Appx, you do it more exact.
So rateH2= sqrt 10 *rate He appx....
so, the time for H2 will be about 1/sqrt10 times rate Ne, or about 44 sec to effuse.
It may be that you are using 140 as the rate, when in fact the rate is the INVERSE of the time to effuse.
How do u know if it is the inverse or not.
and you are not supposed to do 20/2
we are comparing the rate of neon to C4H10 correct?
Titanium metal can be produced from the reduction of titanium(4) chloride with magnesium metal to form solid titanium and magnesium chloride:write the balance equation for the reaction.
To solve this problem, you need to apply Graham's Law of Effusion, which describes the relationship between the rates of effusion of two gases. The formula for Graham's Law is:
Rate1/Rate2 = √(Molar mass2/Molar mass1)
In this equation, Rate1 and Rate2 represent the rates of effusion of the two gases, and Molar mass1 and Molar mass2 represent the molar masses of the two gases.
In your specific problem, you are given the time it takes for neon (Rate1) to effuse, and you need to find out how long it takes for butane (Rate2) to effuse.
However, before we proceed to solve the problem, it's important to understand the units involved in Graham's Law. The time is typically measured in seconds, and according to the law, the rate is inversely proportional to the square root of the molar mass. Therefore, the law implies that the rate of effusion is measured in units of molar mass per square root of time.
Now, let's solve the problem step by step:
1. Start by identifying the given information:
- Rate1 (neon) = 140 seconds
- Molar mass1 (neon) = 20.18 g/mol (rounded to two decimal places)
- Molar mass2 (butane) = ?
2. We need to find the value of Molar mass2 (butane) so that we can calculate Rate2. Rearranging Graham's Law equation, we have:
Rate2 = Rate1 * √(Molar mass1/Molar mass2)
3. Plug in the given values into the equation:
Rate2 = 140 seconds * √(20.18 g/mol / Molar mass2)
At this point, we don't have the exact value of Molar mass2, so we need to keep it as a variable.
4. Now, let's solve for Molar mass2 by rearranging the equation:
Molar mass2 = 20.18 g/mol / (√[(Rate1/Rate2)^2])
Since Rate1 and Rate2 are inversely proportional to the square root of the molar masses, you can equate them and solve for Molar mass2. The square on the right side of the equation cancels out the square root.
5. Substitute the given values into the equation:
Molar mass2 = 20.18 g/mol / (√[(140 seconds / Rate2)^2])
6. Since the time values are in seconds, you do not need to convert them.
7. Calculate the value of Rate2 by substituting the given values:
Rate2 = (Molar mass 2 of butane / Molar mass 1 of neon) * √(Rate1 of neon / Rate2 of butane)
Rate2 = (Molar mass2 / 20.18 g/mol) * √(140 seconds / Rate2)
Now, this equation can be rearranged to calculate Rate2, but it is not possible to isolate Rate2 explicitly due to its presence on both sides of the equation.
Therefore, an equation solver or numerical method (such as iteration) would be required to find the exact value of Rate2. In this case, you might consider using a calculator or a programming language that supports numerical computations.
Once Rate2 is found, you can plug it back into the equation Rate2 = (Molar mass2 / 20.18 g/mol) * √(140 seconds / Rate2) and solve for Molar mass2.
Please keep in mind that the above explanation assumes you have accurate values for molar masses and rates. If you encounter any discrepancies, it is advisable to recheck the values and the calculations to ensure accuracy.