The average rate of reaction over a given interval can be calculated by taking the difference of concentration on a particular given reactant, and dividing it by the total time. In this case, (1.00 M - 0.655 M)/30 s = 0.0115 M/s, or 0.0115 mol/L-s, and this is the final rate of reaction.
824 g NH3 (1 mol/17 g NH3) (4 mol NO/4mol NH3)
48.47 moles NO
Answer: It is non-spontaneous at all T.
Explanation:
According to Gibb's equation:
= Gibbs free energy = +ve
= enthalpy change = +ve
= entropy change = -ve
T = temperature in Kelvin
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 0, reaction is in equilibrium
Putting in the values:
Reaction is non spontaneous at all temperatures.
Answer:
Explanation:
7) A block of wood
has a length of 10.2
cm, a width of 6 cm
and a height of 4.1
cm. The wood has
a total mass of 179
grams. What is the
volume of the
wood and what is
the density of the
wood?
volume = L XW XH=10.2 X 6 X4.1 =250.92cm^3
DENSITY = M/V = 179gm/250.92 cm^3
=0.713 gm/cm^3
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
