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garri49 [273]
4 years ago
15

An element has three naturally occurring isotopes. ten percent (.10) occurs as 55^x, fifteen percent (.15) occurs as isotope 56^

x, and seventy-five percent (.75) occurs as 57^x. calculate the weighted atomic mass of element x to the nearest tenth
Chemistry
1 answer:
Leviafan [203]4 years ago
5 0
The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.

Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>
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4 NH3 (g) + 7 O2 (g) 4 NO2 (g) + 6 H2O (g)
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Aluminum and oxygen react according to the following equation: 4Al(s) 3O2(g) --&gt; 2Al2O3(s) What mass of Al2O3, in grams, can
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Answer:

8.66 g of Al₂O₃ will be produced

Explanation:

4Al (s) + 3O₂ (g)  → 2Al₂O₃ (s)

This is the reaction.

Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.

4.6 g / 26.98 g/mol = 0.170 moles

Ratio is 4:2.

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0.170 moles of Al, may produce (0.170  .2)/ 4 = 0.085 moles

Let's convert the moles of Al₂O₃ to mass.

0.085 mol . 101.96 g/mol = 8.66 g

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3 years ago
Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
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Answer:

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Explanation:

1. First balance the equation for the synthesis of cryolite:

Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}

2. Find the limiting reagent between the Al_{2}O_{3},NaOH and HF

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}

55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

Al_{2}O_{3}:\frac{131}{1}=131

NaOH:\frac{1385}{6}=231

HF:\frac{2770}{12}=231

The Al_{2}O_{3} is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}

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4 years ago
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