Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer:- 2.92 kJ of heat is released.
Solution:- We have water at 100 degree C and it's going to be cool to 15 degree C.
So, change in temperature, 
= 15 - 100 = -85 degree C
mass of water, m = 8.2 g
specific heat of water, c = 
The equation used for solving this type of problems is:
Let's plug in the values in the equation and solve it for q which is the heat energy:
q = (8.2)(4.184)(-85)
q = -2916.248 J
They want answer in kJ. So, let's convert J to kJ and for this we divide by 1000.
q = -2.92 kJ
Negative sign indicates the heat is released. So, in the above process of coiling of water, 2.92 kJ of heat is released.
Ionic bonding is a type of chemical bonding that involves the electrostatic attraction between oppositely charged ions, or between two atoms with sharply different electronegativities, and is the primary interaction occurring in ionic compounds
Answer:
cerium (iii) sulfate is less soluble
