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vazorg [7]
2 years ago
5

How many protons and electrons does oxygen have? What about hydrogen? Draw a model of each atom if that helps you think about th

e differences between them ?
Chemistry
1 answer:
Tems11 [23]2 years ago
5 0

Answer:

Two electrons fit in the first shell out from the nucleus and eight fit in the second. Every element with more protons than the two of Helium needs to work on shells outside the first one. one you get to ten, you have filled the first two shells.

In a water molecule, oxygen forms one covalent bond with EACH of TWO hydrogen atoms. As a result, the oxygen atom has a stable arrangement of 8 valence electrons. Each hydrogen atom forms only one bond because it needs only two electrons to be stable.

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Which of these correctly describes what happens to the molecules in a BOWL OF ICE CREAM when it melts?
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The answer to the question is D
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At approximately what temperature will 40g of argon gas at 2 atm occupy a volume of 22.4L
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3 years ago
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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>

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3 years ago
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IgorC [24]
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3 years ago
The dimerization of butadiene 2C4H61 g2h C8H121 g2 was studied at 500. K, and the following data were obtained: Time (s) [C4H6]
ivanzaharov [21]

Answer:

Rate law: k[C_4H_6]^2

Integrated Rate Law: \frac{1}{[C_4H_6]}=\frac{1}{[C_4H_6]_0}+kt

k = 1.4 \times 10^{-2}

Explanation:

We can see that the graph of time is linear compared to \frac{1}{[C_4H_6]} and the reaction is second order hence we get the rate law from k[A]^n.

The integrated rate law for second order is \frac{1}{[A]}=\frac{1}{[A]_0} +kt where A is C_4H_6.

The slope of the graph \frac{1}{[C_4H_6]} w.r.t time is equal to k. The slope of the graph from the table is 0.014 which is equal to k.

8 0
3 years ago
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