Similar to illustration Similar to illustration Drawing WS 12/6 PLUS MC NE WS Item No. 1927530000 WS, Terminal marker, 12 x 6 mm, Pitch in mm (P): Weidmueller, Allen-Bradley, white Properties Version WS, Terminal marker, 12 x 6 mm, Pitch in mm (P): Weidmueller, Allen-Bradley, white Order No. 1927530000 Type WS 12/6 PLUS MC NE WS GTIN (EAN) 4032248574209 Qty. 600 ST Height 12 mm Height (inches) inch Width 6 mm Width (inches) inch Net weight g Operating temperature range -40...100 °C Application/manufacturer Weidmueller Allen-Bradley Colour white Halogen No Material Polyamide 66 Material colour according to resistance code 9 Number of markers per combination 1 MultiCard = Terminal marker Operating temperature range -40...100 °C Operating temperature range, max. 100 °C Operating temperature range, min. -40 °C Printed characters neutral UL 94 flammability rating V-2 Width 6 mm Engineering Data [CAD data -- STEP]( Engineering Data [EPLAN, WSCAD, Zuken CADENAS eCATALOG [Cax data]( Product Change Notification [Technical change to MultiCard material - DE]( [Technical change to MultiCard material - EN]( Catalogues [Catalogues in PDF-format](LINK/ ETIM EC000761 ETIM EC000761 ETIM EC000761 ECLASS 27-14-11-37 ECLASS 27-14-11-37 ECLASS 27-14-11-37 ECLASS 27-28-11-01 ECLASS 27-28-11-01 Related productsOnline Catalogue » Werkstatt & Zubehör » Markierer » Klemmenmarkierer » MultiCard » Dekafix » DEK 5/6 » Neutral. DEK 5/6 MC NE WS freshman yearStoichiometry stoichiometry worksheet and key kclo3 kcl o2 how many moles of o2 will be formed from moles of kclo3? mol kclo3 mol o2 mol kclo3 mol o2.. Stoichiometry Worksheet and Key mol KClO 3 mol KClO 3 mol O 2 = molO 2 KCl = mol KClO 3 = mol Fe = mol Fe 2O 3 = = 2 KClO 3 è 2 KCl + 3 O 2 10. How. Search: Stoichiometry Part 1 Worksheet Answers. a) Part 1 1 class period to explore the principles of spectrophotometry Worksheet for Basic Stoichiometry Part 1: Mole ←→ Mass Conversions Convert the following number of moles of chemical into its corresponding mass in grams Show all work, answers are given at the end of the problem There is now a fairly. 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