“If a certain metal is cooked, it will conduct electricity better”
- this is a hypothesis, since the statement is kind of vague...”certain metal”...and is written in an “if-then” format.
The amount of electric current that passes through a conductor is directly proportional to the voltage and is described as V = IR
- laws are usually scientific understanding that we’ve tested so much that we’ve come to a conclusion that can often be written out as a mathematical relationship. Hence, the “V = IR”
Electricity is the flow of electric charge in the form of electrons moving through a circuit
- this is a theory. I would give an explanation but I’m out of time.
Hope this helps
Oil and water it’s components do not seperate when left standing
F- due to the fact that it has one more electron.
Hope this helps and good luck :)
The answer I believe would be D) All of these
<span>54.8 g of MgI2 can be produced.
To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium
Atomic weight of Iodine = 126.90447
Atomic weight of Magnesium = 24.305
Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394
Now determine how many moles of Iodine and Magnesium you have
moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole
moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole
Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent.
So figure out how many moles of magnesium will be consumed by the iodine
0.393997154 mole / 2 = 0.196998577 mole.
This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2
0.196998577 mole * 278.11394 g/mole = 54.78805 g
Round the result to the correct number of significant figures.
54.78805 g = 54.8 g</span>
