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Arada [10]
3 years ago
13

The following Lewis diagram represents the valence electron configuration of a main-group element. This element is in group 2A A

ccording to the octet rule, this element would be expected to form a(n) with a charge of cation anion If X is in period 4, the ion formed has the same electron configuration as the noble gas The symbol for the ion is________
Chemistry
1 answer:
Paraphin [41]3 years ago
4 0

<u>Answer:</u> The symbol of the ion formed is Ca^{2+}

<u>Explanation:</u>

An ion is formed when a neutral atom looses or gains electrons.

  • When an atom looses electrons, it results in the formation of positive ion known as cation.
  • When an atom gains electrons, it results in the formation of negative ion known as anion.

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

The element present in Group 2-A and in period 4 is Calcium (Ca)

Electronic configuration of Ca atom:  1s^22s^22p^63s^23p^64s^2

This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)

The electronic configration of Ca^{2+}\text{ ion}=1s^22s^22p^63s^23p^6

Hence, the symbol of the ion formed is Ca^{2+}

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kogti [31]

Answer:

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△H=−72 kcal

The energy required for production of 1.6 g of glucose is [molecular mass of glucose is 180 gm]

b.

\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}

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3 years ago
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I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

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ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

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2 years ago
Convert -12.00 degrees Celsius to the unit Kelvin. 296.15 K 261.15 K -285.15 K -162.85 K
Gre4nikov [31]

To convert Celsius to kelvin, we need to know the formula. Kelvin = Celsius + 273.15. Now, we can figure out how many degrees kelvin -12 degrees Celsius is.


Kelvin = -12 + 273.15

Kelvin = 261.15

7 0
3 years ago
What occurs along a convergent plate boundary ?
Margaret [11]

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Along a convergent plate boundary, two plates moves towards each other as the move in the same direction.

This results in different forms of plate interactions depending on the plate types.

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