Formula for Barium Nitrate = Ba(NO3)2
Thus based on stoichiometry:
1 mole of Ba(NO3)2 contains 2 moles of NO3-
Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M
Use the relation:
V1M1 = V2M2
V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L
Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.
Answer:
What do <u>YOU</u> think?
Explanation:
This question is asking for an opinion. The word should is included.
Explanation:
For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count.
a. 20.0 meters x 0012.65 meters
The numbers are;
20.0 (3 s.f) and 12.65 (4 s.f)
The multiplication gives; 253
Since 253 is already in 3 s.f, that's the answer.
b. 002.5 × 103 meters + 3.50 × 102 meters
The numbers are;
002.5 (2 s.f), 103 (3 s.f), 3.50 (3 s.f) and 102 (3 s.f)
002.5 × 103 = 257.5 = 260 (2 s.f)
3.50 × 102 = 357 = 360 (2 s.f)
260 + 360 = 620
