Answer:
Any of the answers given will work
Explanation:
I literally just did it.
Answer:
Zero
Explanation:
Recall that;
E = q + w
Where;
q = heat, w = work done
When heat is absorbed by the system q is positive
When heat is evolved by the system q is negative
When the system does work, w is negative
When work is done on the system w is positive
Step 1
ΔE1= 60 KJ + 40 KJ = 100KJ
Step 2
ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ
ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ
Depends on where the object is. On earth, moon , or somewhere without any other mass (theoretically). I think you mean how much does weigh on earth. So, the average gravitational acceleration on earth is : 9.83 m/s^2 To find out how much an object weighs, this is the formula : G=m.g where m is mass of the object g is the gravitational acceleration and G is weight. So, G = 10.9,83 = 98,3 N is the answer.
Answer: yes
Explanation:
Not all the time, but temperature change can happen during a chemical reaction.
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
