Answer:- 14.0 moles of hydrogen present in 2.00 moles of ![[tex]C_6H_7N](https://tex.z-dn.net/?f=%20%5Btex%5DC_6H_7N)
.
Solution:- We have been given with 2.00 moles of 
and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:
= 14.0 g H
So, there are 14.0 g of hydrogen in 2.00 moles of .
<span>0.70 mol/0.250 L = 2.8 M</span>
Reduction <span>always results in a lowering of the oxidation number. The reaction of the system above is written as:
</span><span>Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq)
</span>
From the reaction, we see that copper goes from the +2 to a neutral charge. Lowering of the oxidation number happens so this is the element that is being reduced.
Answer:
The sum of the molar masses of each isotope of the element.
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
