Answer:
C
) 2, 1, 2
Explanation:
The given reaction is synthesis reaction in which lithium and bromine react to form lithium bromide.
Chemical equation:
Li + Br₂ → LiBr
Balanced chemical equation:
2Li + Br₂ → 2LiBr
Step 1:
Li + Br₂ → LiBr
left hand side Right hand side
Li = 1 Li = 1
Br = 2 Br = 1
Step 2:
Li + Br₂ → 2LiBr
left hand side Right hand side
Li = 1 Li = 2
Br = 2 Br = 2
Step 3:
2Li + Br₂ → 2LiBr
left hand side Right hand side
Li = 2 Li = 2
Br = 2 Br = 2
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:
the answer is b Li + Cl2 .....
Answer:
8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days
Explanation:
The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.
We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.
Initial amount of Thorium-234 available as per the question is 150 grams
So now we start with 150 grams of Thorium-234
So after 120.5 days the amount of sample that remains is 8.625g
In simpler way , we can use the below formula to find the sample left
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
