Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require = 1 mole of 
Thus 0.061 moles of will require=
of 
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 1 mole of 
Thus 0.061 moles of give = of
Mass of 
Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.
Answer:
Animal was shot in Nunavut, Canada earlier this month by hunters. It resembles a polar bear but has claws and brown paws of a grizzly bear. Experts claim the 'pizzly' or 'grolar' bear is a hybrid between the two. DNA tests revealed it was actually a blonde haired grizzly bear
Explanation:
Answer:
1) Increase temperature
2) Decrease temperature
3) Increase concentration of reactants
4) Increase pressure
5) Decrease pressure
Explanation:
Le Chatelier's Principle Fundamentals states that a chemical reaction at equilibrium that undergoes changes to pressure, temperature, or concentration, this will cause the equilibrium to shift in the opposite direction to offset the change.
1) Increase temperature
2) Decrease temperature
3) Increase concentration of reactants
4) Increase pressure
5) Decrease pressure
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
