<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>
<span>When storing parts outside, it is best to keep them under tarps or wrapped to stay in compliance?
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15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
The density of ice is less than the density of water.
Fun fact: Water is the only known liquid that has this change it density.
Fifty percent. plz brainliest my answer
