She will cover distance in 3 days and 16 hours .the time will be 3:15 am on the fourth day.
Convert 5.9 miles/hour to feet/hour there are 5,280 feet in a mile
total distance covered = 47500
distance = rate x time
now, time = 47500/5280
time = 88 hours
88 hours = 3 Days and 16 Hours.
now the time will be 3 :15 am on the fourth day.
The overall distance the object covers in a given amount of time is its average speed. A scalar value represents the average speed. It has no direction and is indicated by the magnitude.
The distance traveled divided by the time taken is the most popular formula for calculating average speed. The alternative formula is to add the initial and final speeds together, then divide by 2.
Learn more about distance time problems here: -
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Answer:
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Explanation:
You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
Active transport is the moving of molecules across the membrane of the cell against the concentration gradient with the use of ATP.
Low to high concentration. Concentration gradient is the diffusion (movement of molecules from regions of low concentration) from high to low with the gradient. Active transport is from low to high, against the gradient.
