Atoy train is pushed forward and released at 2.0m with a speed of 1.0m/s. it rolls at a steady speed for 2.0s, then one wheel begins to stick. the train comes to a stop at 5.0m from the point at which it was released. what is the train's acceleration after its wheel begins to stick?

Recall that

where

- is the final velocity of the train; 0 in this case, because the train eventually stops

- is the initial velocity of the train; 1.0 m/s in this case, because this initial velocity refers to velocity it starts with after one of the wheels gets stuck, and the train is initially traveling at a constant speed of 1.0 m/s

- is the acceleration we want to find

- is the change in the train's position; 5.0 m in this case

So we have

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

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