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lidiya [134]
4 years ago
15

The carbon-14 radioisotope has proved most useful in:

Chemistry
1 answer:
valkas [14]4 years ago
8 0
Measuring the amount of radiation in an area
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Which terms are used to identify pure substances
Semenov [28]

Answer:

Color, odor, density, melting temperature, boiling temperature, and solubility are examples of physical properties. Physical properties can be used to identify a pure substance.

Explanation:

please mark brainliest! :)

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Convert 53.4 g to ounces
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53.4 grams is converted into 1.88363 ounces
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Round 54,878 to the nearest ten thousand
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55,000 because 878 is bigger than 499. Also 54 is in the place where 10 is in 10,000.

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The outer planets, referred to as the gas giants, contain and satellites.
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In pure water at 25 °C, the concentration of a saturated solution of CuF2 is 7.4 × 10−3 M. If measured at the same temperature,
Romashka-Z-Leto [24]

Answer:

The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.

Explanation:

Consider the ICE take for the solubility of the solid, CuF₂ as:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -              -

At t =equilibrium      (x-s)                s           2s          

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

K_{sp}=s\times {2s}^2

K_{sp}=4s^3

Given  s = 7.4×10⁻³ M

So, Ksp is:

K_{sp}=4\times (7.4\times 10^{-3})^3

K_{sp}=4\times (7.4\times 10^{-3})^3

Ksp = 1.6209×10⁻⁶

Now, we have to calculate the solubility of CuF₂ in NaF.

Thus, NaF already contain 0.20 M F⁻ ions

Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:

                                  CuF₂    ⇄     Cu²⁺ +    2F⁻

At t=0                            x                 -            0.20

At t =equilibrium      (x-s')             s'         0.20+2s'         

The expression for Solubility product for CuF₂ is:

K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2

1.6209\times 10^{-6}={s}'\times ({0.20+2{s}'})^2

Solving for s', we get

<u>s' = 4.0×10⁻⁵ M</u>

<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is  4.0×10⁻⁵ M.</u>

3 0
4 years ago
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