I'm horrible at chemistry. it just bothers me. right now we are learning the problem solving steps. and it is just annoying to me, although i've finally starting to get it. Although these couple problems are just stumping me.A 100 gram sample of iron ore was round to contain 44 g of iron. How many grams of iron are in a 450 gram sample of ore?the density of a liquid is .821g/mL. What is the volume of 71.3 g of this liquid?The density of lead is 11.3 g/cm cubed. What is the mass of a cube of lead 10.0 cme on each edge?thanks to everyone who attempts to help me!!!! :D
depends on the car. some cars have them. call the dearler, they will tell you.
not on all cars one way to test it is to lock your self in and open the door
no, my car has a remote to unlock and lock the doors, it even chirps like most alarms, but it has no alarm in it at all. I have tested it and there is no alarm, I also did not pay the $1000 extra for the alarm.
A 100 gram sample of iron ore was found to contain 44 g of iron. How many grams of iron are in a 450 gram sample of ore? Just set up a proportion: (44 g iron) / (100 g ore) (x g iron) / (450 g ore) Now solve for x: x 198 g of iron ------------------------------------ The density of a liquid is .821 g/mL. What is the volume of 71.3 g of this liquid? Use the units to help guide your calculations. We wish to have the answer in the form of volume units, which in this case is mL: (71.3 g) / (0.821 g/mL) 86.8 mL Notice how the g units cancel, leaving the unit 1/(1/mL), which equals mL. Therefore, not only do the units tell us to divide, they tell us how to divide. ------------------------------------ The density of lead is 11.3 g/cm?. What is the mass of a cube of lead 10.0 cm on each edge? Again, use the units to help guide your calculations (notice how I rewrote the volume unit above). We wish to have the answer in the form of mass units, which in this case is g. Also, we know that the volume of a cube equals the cube of the length of its edge. Therefore: (11.3 g/cm?) (10 cm)? 11,300 g In this case, the units have told us to multiply the density times the cube of the edge, so that the cm units will cancel. We really didn't have to know the volume of a cube formula, the units told us to cube the 10 cm.
