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Lady bird [3.3K]
3 years ago
5

How many protons, electrons, and neutrons does the following isotope contain? ^{18} \text{F}^{-} 18 F −

Chemistry
1 answer:
melomori [17]3 years ago
8 0
Isotope ¹⁸F⁻ contains:
1) p⁺ = 9; number of protons.
Fluorine has a<span>tomic number Z = 9 (total number of protons).
2) e</span>⁻<span> = 10; </span>number of electrons.<span>
In element number of electrons and protons are the same, because element has neutral charge, but because in this example, fluorine is anion with negative charge, it has one electron more.
3) n</span>° = 9; number of neutrons.
<span>Mass number A = 18 is total number of protons and neutrons in a nucleus, so number of neutrons is A-Z = 18-9=9.</span>
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B

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For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
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5 0
2 years ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

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Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0
3 years ago
Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.89×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.25×10−4, what is the
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Answer:

1.2\times 10^{-2}

Explanation:

The given reactions are:

PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq)     K_3 = 1.89\times 10^{-10}

AgCl(aq)⇌Ag+(aq)+Cl−(aq)           K_4 = 1.25\times 10^{-4}

Required reaction is:

PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

K_f = \frac{K_3}{K_4^2}\\=\frac{1.89\times 10^{-10}}{1.25\times 10^{-4}}^2\\=1.2\times 10^{-2}

4 0
3 years ago
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