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
<span>Persamaan setara pada reaksi besi dengan asam klorida membentuk besi (II) klorida dan gas hidrogen
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Answer:
D) winds that blow in the same direction at a consistent speed
Explanation:
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