Answer:
The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = Heat gained = ?
c = Specific heat = 
ΔT = The change in temperature = 3.08°C
Now put all the given values in the above formula, we get:
Now we have to calculate molar enthalpy of combustion of this substance :
where,
= enthalpy change = ?
q = heat gained = 8.2544kJ
n = number of moles methane = 
Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
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
Mass number = protons + neutrons = 34+46 =80
the element with an atomic number of 34 and mass number of 80 is selenium
Explanation:
It is given that,
The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :
The radius of the orbit in which time period is 8T is R'. So, the relation is given by :
So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.
A good example is the mineral<span> plagioclase. Plagioclase is a member of the feldspar group, but </span>there<span> is more than one type of plagioclase.</span>
