Answer:
0.0428 M
Explanation:
Because we're asked to calculate the molarity of nickel(II) cation, we need to <u>determine all sources for that species</u>, in this case, all Ni⁺² comes from the nickel(II) bromide solid (NiBr₂).
We use the molecular weight of NiBr₂ to calculate the moles of Ni:
1.87 g NiBr₂ ÷ 218.49g/mol * (1molNi⁺²/1molNiBr₂) = 8.55x10⁻³ mol Ni⁺²
Then we <u>divide the moles by the volume in order to calculate the concentration</u>:
8.55x10⁻³ mol Ni⁺² / 0.200 L = 0.0428 M
Answer:
7
Explanation:
(Note: for a strong acid and strong base titration the equivalence point is at a pH=7. This is because at this point you have equal moles of added base as acid in the original solution. Therefore at the equivalence point the solution has formed a neutral salt and the pH is 7).
Answer:
Mass of iron = 59.375 gm
Explanation:
Calories ( or joules) are added to the water by the hot steel so at the endpoint they are BOTH at 40 C
The water gains:
4.18 j/g-C * 50 * (40-30 C) = 2090 j
The steel gave up 2090 j going from 120 to 40 C
2090 = .44 j/g-C * m * (120-40) solve fro m = 59.375 gm
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