<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>
Water is found in lakes and rivers. Its purpose in nature is to hydrate animals
Hydrogen gas is not abundant in nature, but hydrogen is abundant in water. Hydrogen (bonds) help ice float. Otherwise, freezing lakes would kill animals (it actually doesn't since the ice creates a "coat" above the water").
Ammonia is nitrogen-rich molecules that plant uses to get their nitrogen. It comes from the part of the nitrogen cycle where dead plants and animals are decayed.
Carbon dioxide is what we exhale. Plants "inhale" carbon dioxide and use that for photosynthesis.
Hydrogen sulfide is emitted by volcanoes and by anaerobic (oxygen-less) decay from bacteria.
Have an awesome day! I hope this helps.
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
0.444 mol/L
Explanation:
First step is to find the number of moles of oxalic acid.
n(oxalic acid) = 
Now use the molar ratio to find how many moles of NaOH would be required to neutralize 
of oxalic acid.
n(oxalic acid): n(potassium hydroxide)
1 : 2 (we get this from the balanced equation)
: x
x = 0.0111 mol
Now to calculate what concentration of KOH that would be in 25 mL of water:
I think your answer is A not sure tho
