This problem is providing us with the mass equivalent to one troy ounce. Thus, the troy ounces of gold in one short ton of average Nevada ore is required and found to be the 0.103 otz according to the following dimensional analysis.

<h3>Dimensional analysis</h3>

In chemistry, a raft of problems do not always provide an equation in order to be solved yet dimensional analysis can be applied, so as to obtain the desired amount in the required units.

Thus, since this problem asks for try ounces in an average Nevada ore, which has 3.2 grams of gold per short ton of ore, one can solve the following setup in order to obtain the required answer in otz:

$\frac{3.2gAu}{1short-ton}*1short-ton*\frac{1otz}{31.1g} \\$

Where the short tons are cancelled out as well as the grams, in order to obtain:

$0.103 otz$

Learn more about dimensional analysis: brainly.com/question/10874167

4 0

2 years ago

Unstable isotopes undergo radioactive decay. what occurs during radioactive decay?

mel-nik [20]

Here is what radioactive decay is:
<span>Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.</span>

5 0

3 years ago

How does the volume of water change the solubility of sodium chloride in simple words?

kupik [55]

Answer:

When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. ... The solubility of salt does not change much with temperature, so there is little profit in using hot water.

3 0

3 years ago

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

Taya2010 [7]

Answer:

Here's what I get

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial points

The graph below shows the initial concentrations plotted on the vertical axis.

7 0

3 years ago