Answer:
39.2 g
Explanation:
- 2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)
First we <u>convert 55.3 grams of Ni₂O₃ into moles of Ni₂O₃</u>, using its<em> molar mass</em>:
- 55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃
Then we <u>convert 0.334 moles of Ni₂O₃ into moles of Ni</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.334 mol Ni₂O₃ *
= 0.668 mol Ni
Finally we <u>calculate how much do 0.668 Ni moles weigh</u>, using the<em> molar mass of Ni </em>:
- 0.668 mol Ni * 58.69 g/mol = 39.2 g
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The balanced chemical equation for the production of chromium metal from the reaction of chromium(ll) nitrate reacts with a strip of zinc is:
3 Zn + 2 Cr(NO₃)₃ → 2 Cr + 3 Zn(NO₃)₂
This is a redox reaction, which <u>is a chemical reaction in which one or more electrons are transferred between the reagents</u>, causing a change in their oxidation states. In the proposed reaction, Cr oxidation state goes from +3 to 0, becoming metallic chromium, while Zn goes from being Zn⁰ to Zn²⁺.
<u>The mass of chromium metal produced in the above reaction will be,</u>
425.0 mL x
x
x
x
= 5.52 g
So, the mass of chromium metal produced when 425.0mL of 0.25M chromium(ll) nitrate reacts with a strip of zinc that remains in excess is 5.52 g of Cr.
Answer:
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Explanation: