Answer:
The answer to your question is 35.4527 g
Explanation:
Data
Cl-35 abundance = 75.77 % mass Cl-35 = 34.9688
Cl-37 abundance = 24.23 % mass Cl-37 = 36.9659
Process
1.- Convert the percents to decimals
75.77% = 75.77/100 = 0.7577
24.23% = 24.23/100 = 0.2423
2.- Find the atomic weight
Average atomic weight = Abundance 1 x Mass Cl-35 +
Abundance 2 x Mass Cl-37
Substitution
Average atomic weight = (0.7577)(34.9688) + (0.2423 x 36.9659)
Simplification
Average atomic weight = 26.4959 + 8.9568
Result
Average atomic weight = 35.4527 g
The correct answer for, An element with the smallest anionic (negative-ionic) radius would be found on the periodic table in, is <span>Group 17, Period 2.</span>
Answer:
51.0 meters/second is the final speed
Simply atomic masses are to be added since the formula is KCl. The atomic mass of KCl is 74.6 u , and its molar mass is 74.6 g/mol
Carbon
calcuim
iron
hydrogen
iodine
nitrogen
oxygen
phosphorus
sulfur
The letters are the symbols for those elements
