Answer:
25 grams
Explanation:
You strat off with 400 grams of your substance. By day 14, half has dcayed and you only have 200 grams left. By day 28, there are 100 grams of the substance. On day 42, there are 50 grams left. Finally, on day 56, the substance has been through four half-lives and 25 grams remain.
Answer:
v = 23.96 cm³
Explanation:
Given data:
Mass = 15.0 g
Density = 0.626 g/cm³
Volume = ?
Solution:
Formula:
D=m/v
D= density
m=mass
V=volume
Now we will put the values in formula:
d = m/v
v = m/d
v = 15 g / 0.626 g/cm³
v = 23.96 cm³
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)
Answer:
A3B3
Explanation:
Molecular formula = n x empirical formula
(AB) n = 90
MM of AB = 30 g/mol
30n = 90
Divide both side by the coefficient n i.e 30
n = 90/30 = 3
Molecular formula = n x empirical formula
Molecular formula = n x (AB)
Molecular formula = 3(AB) = A3B3
