<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>
The answer Is (1)...Straight Isomer when you put them together.
Hope that helps!!!
NaHCO3 is a product of a strong base and a weak acid reaction. Thus it has weak basic properties.
HCO3- ion is actually amphoteric, which means it can act as a base or an acid. But it is weaker than a strong acid or a strong base.
<span>HCO3- is amphoteric meaning it acts both as a B.L. Acid and a B.L. Base.. which is why it's used to neutralize both acid and base spills in the lab.</span>
Answer:
Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).
