What should you do after fueling your boat and before starting the engine?

After Fueling of the Engine of your Boat you should proper check that here is no other vapours of the Fuel in the engine room of the boat.

Further Explanation:

Tips after fueling of engine of boat:

There are 5 main tips which you can do after the fueling of the engine.

First step:

You have to close the lid of the engine of the boat so that there are no process of evaporation. And the vapors are not evaporated.

Second Step:

Open all the doors and windows of your boats so that there are no vapors left.

Third Step:

If you have any exhaust system then on it for some minutes this will help you to remove all vopurs of fuel from the fuelling room of  the engine.

Fourth Step:

After all this process sniff all the engine parts to ensure that there are no vapors left in the engine room

Starting of engine :

After doing all these steps you can start the engine with no fear. Your engine will starts with full zeal and zest.

Answer Details:

Subject: Physics

Level: High School.

Key Words:

Tips after fueling of engine of boat:

First step:

Second Step:

Third Step:

Fourth Step:

Starting of engine :

For further Evaluation:

thanks

  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

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.