0.400✓STARTING AMOUNT0.400 L X0.350 mol KBr1 LX16.7 g KBr0.350mol KBr✓What is the mass in grams of KBr in 0.400=ADD FACTORx( )X0.4000.14016.75.4716.7mol KBrg KBrmLDELETEXA39.1016.022 × 10231.18 × 10-³M KBrLL of a 0.350 M solution?ANSWER0.3501367.09 × 102⁰9.60 × 10-³g KBrRESET38.43 × 10²²g KBr/mol119

Answered: Image /qna-images/answer/ca7a1b10-fdd1-45cf-a7a4-66cbde105dae.jpg

X STARTING AMOUNT 0.400 0.400 L x 0.350 mol KBr 1 L X = 16.7 g KBr 0.350 1 mol KBr X What is the mass in grams of KBr in 0.400 L of a 0.350 M solution? = 16.7 ADD FACTOR 0.400 0.140 5.47 16.7 mol KBr g KBr DELETE mL ㅗㅜ 39.10 6.022 x 1023 1 1.18 x 10³ M KBr L ANSWER 0.350 136 9.60 × 10³ RESET 7.09 x 1020 8.43 x 10²² g KBr 2 g KBr/mol 119

X STARTING AMOUNT 0.400 0.400 L x 0.350 mol KBr 1 L X = 16.7 g KBr 0.350 1 mol KBr X What is the mass in grams of KBr in 0.400 L of a 0.350 M solution? = 16.7 ADD FACTOR 0.400 0.140 5.47 16.7 mol KBr g KBr DELETE mL ㅗㅜ 39.10 6.022 x 1023 1 1.18 x 10³ M KBr L ANSWER 0.350 136 9.60 × 10³ RESET 7.09 x 1020 8.43 x 10²² g KBr 2 g KBr/mol 119

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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