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sdas [7]
3 years ago
14

What is the charge in coulombs on an electron?

Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The charge on an electron is ~

  • \sf{ - 1.602 \times 10 {}^{ - 19}  \:  \: Coulombs}

" - " symbol is used to show that its negative.

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(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most
Charra [1.4K]

Answer:

(a) The Lewis structure is shown in the image below.

(b) CaF_2

(c) Calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond.

(d) Calcium atom loses electrons.

Explanation:

Calcium is the element of second group and forth period. The electronic configuration of Calcium is - 2, 8, 8, 2 or 1s^22s^22p^63s^23p^64s^2

There are 2 valence electrons of Calcium.

Fluorine is the element of group 17 and second period. The electronic configuration of the element fluorine is - 2, 7 or  1s^22s^22p^5

There is 1 valence electron of fluorine.

The Lewis structure is drawn in such a way that the octet of each atom is complete.  

Thus, calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Thus, the formula of calcium chloride is CaF_2.

<u>(a) The Lewis structure is shown in the image below.</u>

<u>(b) CaF_2</u>

<u>(c) Calcium loses 2 electrons to 2 atoms of fluorine and these 2 atoms of fluorine accepts each electron to form ionic bond.</u>

<u>(d) Calcium atom loses electrons.</u>

5 0
3 years ago
Consider the reaction: 2 H2O (g)--&gt;2 H2 (g) + O2 (g). ΔH=483.6 Kj/mol. If 2 moles of H2O (g) are converted H2(g) and O2(g) ag
denis-greek [22]
DE = dH - PdV 

<span>2 H2O(g) → 2 H2(g) + O2(g) </span>

<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>

<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>

<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>

<span>2 moles reacts to form 3 moles </span>

<span>The gas equation is </span>

<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
<span>R = gas constant = 8.314 J K^-1 mol^-1 </span>
<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>

<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>

<span>dE = dH -dnRT (because PV = nRT) </span>

<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>

<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>

<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>
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