Releasing the string would most likely increase the kinetic energy, as kinetic energy is energy in motion. However, if you were to increase the potential energy, then the string would have to be pulled back more, as potential energy is stored energy within an object.
The number of solid precipitate that will be formed is 37.08 g
calculation
write the equation for reaction
=Hg(NO3)2 +Na2SO4 = HgSO4(s) +2NaNO3(aq)
find the moles of each reactant
moles ofHg(NO3)2=126.27/324.6= 0.389 moles
moles of Na2SO4=17.796/142=0.125 moles
NaSO4 is the limiting reagent and by use of mole ratio of NaSO4:HgSO4 which is 1:1 therefore the moles of H2SO4 is also= 0.125 moles
mass HgSO4=moles x molar mass
=0.125 x296.65= 37.08g
I think it’s a, it has to be
Answer:
Final number of moles = 0.675 mol
Mass = 2.7 g
Explanation:
Given data:
Initial volume of gas = 2.00 L
Final volume of gas = 2.70 L
Initial number of moles = 0.500 mol
Final number of moles = ?
Solution:
Formula:
V₁/n₁ = V₂/n₂
V₁ = Initial volume
n₁ = Initial number of moles
V₂ = Final volume of gas
n₂ = Final number of moles
Now we will put the values in formula.
2.00 L /0.500 mol = 2.70 L / n₂
n₂ = 2.70 L× 0.500 mol /2.00 L
n₂ = 1.35 L.mol / 2.00 L
n₂ = 0.675 mol
B)
Grams of helium added = ?
Solution:
Mass = number of moles × molar mass
Mass = 0.675 mol× 4 g/mol
Mass = 2.7 g