How do I round 7,366.3275 to an exponent. My answer is 7.37 x 10 to 7th power am I right.
Thanks
6 answers
No. When you say to round you must also say to how many places. If you want it rounded to 3 places AND in exponential form, it is 7.37E3
If the standard atmosphere (1atm) is equal to 1.013 25 X 10fifth exponent. Then How do I put 7,66.3257 to expotent notation. Thanks
Is that 7,366.3257 in what units? Torr? Do you want to convert that to Pascals or atmospheres? How many significant figures do you want?
By the way, most internet forums, such as this one, do not treat subscripts and superscripts very well but the way you're doing it is just too much work.
Use E notation or ^ notation. For example, you might write 1.01325E5 or 1.01325 x 10^5. The E notation is much easier.
Another by the way, 1 atm = 1.01325E5 (but when you don't show the units) it makes no sense. I assume you mean 1.01325E5 kPa because 1 atm = 101.325 kPa.
By the way, most internet forums, such as this one, do not treat subscripts and superscripts very well but the way you're doing it is just too much work.
Use E notation or ^ notation. For example, you might write 1.01325E5 or 1.01325 x 10^5. The E notation is much easier.
Another by the way, 1 atm = 1.01325E5 (but when you don't show the units) it makes no sense. I assume you mean 1.01325E5 kPa because 1 atm = 101.325 kPa.
Thanks
Given the following:
4Al+ 3O2 ==> 2Al2O3
If 3.17g of Al and 2.55g of O2 are available, which reactant is limiting.
My answer is: 1.59 mol Al2O3 and 1.70 mol Al2O3. Al is limiting.
Is my answer and work correct?
4Al+ 3O2 ==> 2Al2O3
If 3.17g of Al and 2.55g of O2 are available, which reactant is limiting.
My answer is: 1.59 mol Al2O3 and 1.70 mol Al2O3. Al is limiting.
Is my answer and work correct?
See your post above. Al is limiting but your calculation is faulty. Damon worked the problem for you correctly.