According to
Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".
r₁ / r₂ =
---- (1)
r₁ = Rate of effusion of He
r₂ = Rate of Effusion of O₃
M₁ = Molecular Mass of He = 4 g/mol
M₂ = Molecular Mass of O₃ = 48 g/mol
Putting values in eq. 1,
r₁ / r₂ =
r₁ / r₂ =
r₁ / r₂ =
3.46
Result: Therefore, Helium will effuse
3.46 times more faster than Ozone.
2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O
<u>Explanation:</u>
C₇H₆O₂ + O₂ → CO₂ + H₂O
First we have to balance the O- atoms, we have to put 6 in front of water so there are 12 H atoms on RHS, to balance it we need to put 2 in front of C₇H₆O₂, and so we have 14 C - atoms on LHS, 28 + 6 = 34 O - atoms on RHS, so we have to put 15 in front of Oxygen in LHS, so that each and every atom in the equation gets balanced now. The balanced equation is, 2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O
Good grief, this stuff got caught in a black hole somewhere. It is terribly dense.
1 mL = 1 cc under normal conditions.
d = mass / volume
m = 20 kg
v = 5 mL
d = 20kg / 5 mL
d = 4 kg / mL
d = 4 kg / cc
A <<<<answer
Answer:
distance between them , and mass of the objects
Explanation:
the further away a object is the weaker the gyrational pull is, and imagine lifting a anvil above your head, the anvil weight push's down upon you pulling.
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate = 
Moles of sodium dichromate = 
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.
