A train with an initial velocity of 31 m/s begins accelerating at rate of 0.0705 m/s^2. If the train travels for 180.5s, how far does it travel?

v = 43.72 m/s

Explanation:

Given that,

Initial velocity of the train, u = 31 m/s

Acceleration of the train, a = 0.0705 m/s²

Time for which the train travel, t = 180.5 s

We need to find the final velocity of the train. Let it is v.. Using first equation of kinematics to find it such that,

So the final speed of the train is equal to 43.72m/s.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"