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natima [27]
4 years ago
9

What is the maximum mass of pure gold that could be extracted from 3.0kg of calaverite, a gold ore with the chemical formula AuT

e2? Be sure your answer has a unit symbol, if necessary, and is rounded to 2 significant digits.
Chemistry
1 answer:
Marina86 [1]4 years ago
3 0

Answer:

1.3 × 10³ g

Explanation:

Step 1: Convert the mass of calaverite to grams

We will use the relationship 1 kg = 1,000 g.

3.0kg \times \frac{1,000g}{1kg} = 3.0 \times 10^{3} g

Step 2: Establish the appropriate mass ratio

The mass ratio of AuTe₂ to Au is 452.17:196.97.

Step 3: Calculate the maximum mass of pure gold that can be obtained from 3.0 × 10³ g of calaverite

3.0 \times 10^{3} g Calaverite \times \frac{196.97gAu}{452.17gCalaverite} = 1.3 \times 10^{3} gAu

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What is the frequency and energy per quantum (in Joules) of :
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(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

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<h3>Explanation</h3>

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at 3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1} in vacuum.

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(a)

Convert all units to standard ones.

\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}.

The unit of f shall also be standard.

f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}.

For each particle,

E = h\cdot f,

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E = h \cdot f = 6.63\times10^{-34}\;\text{J}\cdot\text{s}\times 5.00\times 10^{20}\;\text{s}^{-1} = 3.32\times10^{-13}\;\text{J}.

(b)

Try the steps in (a) for this beam of microwave with

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Expect the following results:

  • f = 1.20\times 10^{10}\;\text{Hz}, and
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