Answer:
B₂
Explanation:
The limiting reactant is always a reactant. You can determine which reactant is limiting by identifying which has the smaller mole-to-mole ratio with the product. This ratio can be found via the coefficients of the balanced reaction.
4 A₂ + 3 B₂ ---> 6 AB
4 moles A₂
------------------ = mole-to-mole ratio A₂/AB
6 moles AB
3 moles B₂
------------------ = mole-to-mole ratio B₂/AB
6 moles AB
Since the mole-to-mole ratio between B₂ and AB is smaller, B₂ must be the limiting reactant.
Answer:
50/36 = 25/18
Explanation:
Solution at attachment box
Molality = mole of dissolvable (this question glucose) / kg of water
We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
<em> XeF₆</em>
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