<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>
An occluded front is well known to bring a large amount of rain and snow (Option F).
<h3>What is an occluded front?</h3>
An occluded front is a mass of air capable of bringing precipitation in the form of liquid water (rain) or snow.
This meteorological phenomenon (occluded front) is formed by warm air trapped with cold air.
In conclusion, an occluded front is well known to bring a large amount of rain and snow (Option F).
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14.285 % is the answer maybe but I am not sure
I think it’s Object 3
Hopefully this helps
The Molar concentration of your analyte solution is 1.17 m
<h3>What is titration reaction?</h3>
- Titration is a chemical analysis procedure that determines the amount of a sample's ingredient by adding a precisely known amount of another substance to the measured sample, with which the desired constituent reacts in a specific, known proportion.
Make use of the titration formula.
The formula is molarity (M) of the acid x volume (V) of the acid = molarity (M) of the base x volume (V) of the base.
if the titrant and analyte have a 1:1 mole ratio. (Molarity is a measure of a solution's concentration represented as the number of moles of solute per litre of solution.)
26 x 1.8 = 40 x M
M = 26 x1.8 /40
M = 1.17
The Molar concentration of your analyte solution is 1.17 m
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