The correct answer is Oxyanion Nitrite. Oxyanion is referred to as an ion which contains one or more oxygen atoms that are bonded to another chemical element. It has a generic formula of <span>A. </span>x<span>O </span>z− <span>y, wherein A, stands for a chemical element, while O, stands for an Oxygen atom</span>
Answer:
The answer to your question is 7.4 moles of Aluminum
Explanation:
Data
moles of Al = ?
moles of Al₂O₃ = 3.7
Balanced chemical reaction
4 Al + 3 O₂ ⇒ 2 Al₂O₃
To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.
4 moles of Aluminum ----------------- 2 moles of Al₂O₃
x ----------------- 3.7 moles of Al₂O₃
x = (3.7 x 4) / 2
x = 14.8 / 2
x = 7.4 moles of Aluminum
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Hope this helps you :)
Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
Answer:
5.7 moles of O2
Explanation:
We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:
2KClO3 —> 2KCl + 3O2
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.
Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:
From the balanced equation above,
2 moles of KClO3 decomposed to produce 3 moles of O2.
Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.
Thus, 5.7 moles of O2 were obtained from the reaction.
