As,
5471 kJ heat is given by = 1 mole of Octane
Then,
5310 kJ heat will be given by = X moles of Octane
Solving for X,
X = (5310 kJ × 1 mol) ÷ 5471 kJ
X = 0.970 moles of Ocatne
So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
Moles = mass / M.mass
Or,
Mass = Moles × M.mass
Putting values,
Mass = 0.970 mol × 114.23 g/mol
Mass = 110.83 g of Octane
0.4 g/ml.............................
Silver (Ag) is the number of atoms per unit cell for each metal. Silver has a face-centred cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.
The identical unit cells are defined in such a way that they take up space without touching one another. A crystal's internal 3D arrangement of atoms, molecules, or ions is known as its lattice. It consists of a large number of unit cells. Every point of the lattice is occupied by one of the three component particles.
Primitive cubic, body-centred cubic (BCC), and face-centred cubic are the three types of unit cells (FCC). The three different sorts of unit cells will be thoroughly covered in this section.
To learn more about the unit cell refer here:
brainly.com/question/13433017
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Answer:
The molecular formula of the compound :
Explanation:
The empirical formula of the compound =
The molecular formula of the compound =
The equation used to calculate the valency is :
We are given:
Mass of molecular formula = 86 g/mol
Mass of empirical formula = 43 g/mol
Putting values in above equation, we get:
The molecular formula of the compound :
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
