FAQs on Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant
How much does "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" cost at Strange Adventures Tempe?
The price of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" at Strange Adventures Tempe is USD $49.99.
What brand is "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" associated with?
"Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" is associated with the Batman brand.
Who is the publisher of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant"?
The publisher of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" is DC Entertainment.
What is the genre of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant"?
The genre of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" is Superheroes.
What category does "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" fall under?
"Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" falls under the category of Comics.
Who is the artist and cover artist of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant"?
The artist of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" is Eduardo Risso. The cover artist is Tim Sale.
Who is the writer of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant"?
The writer of "Batman the Long Halloween: The Last Halloween #1 (Of 10) Cover E 1 for 50 Incentive Tim Sale Variant" is Jeph Loeb.