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3 years ago

How is scientific use of the term digital different from the common use?

Chemistry

2 answers:

Alenkasestr [34]3 years ago

6 0

Do you want to be a scientist or not because I can help you with that it's so easy

faltersainse [42]3 years ago

6 0

We commonly know refer to something 'digital' has to something electronic that can be visibly seen such as a watch, clock, camera, screen, etc. It really refers to stored energy or electricity that's not natural. But the word 'digital' in science refers to the depiction of data or information in figures (such as in a table) in contrast to as a chart, graph, drawing, or other pictorial form.

Explanation:

The scientific use of the term digital is much changed from the current use because the term digital in science is a way of representing a discontinuous phenomenon or something this is just finite and faulty. It also describes a data in figures with the use of this term. While “digital” generally refers to electronics in general, the scientific explanation of digital is much different

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Translation: Explain the importance of energy levels,

tigry1 [53]

Answer: Formation of a stable Duplet or Octet structure

Explanation:

Energy levels refers to the definite amount of energies that electrons can have when occupying specific orbitals.

Examples are:

Level 1 as in 1s

Level 2 as in 2s, 2p

Level 3 as in 3s 3p 3d

Level 4 as in 4s 4p 4d 4f

Level 5 as in 5s .......

Level 6 as in 6s .........

Electrons can be excited to higher energy levels by absorbing energy from the surroundings during chemical reactions and so forming chemical bonds like ionic, covalent etc.

A perfect example is Sodium (Na) and Chlorine (Cl) in Sodium chloride (NaCl)

Sodium has atomic number of 11, so energy level is 1s2, 2s2 2p6, 3s1

Chlorine has atomic number of 17, so energy level is 1s2, 2s2 2p6, 3s2 3p5

You will notice that the both Na and Cl, has electrons in the energy level 3. Then, the bond formed must be one that lead to the formation of a OCTET structure: as in Na gives off its 3s1 electron TO Cl.

Thus, Cl now becomes 1s2, 2s2 2p6, 3s2 3p6 - an OCTET STRUCTURE (with completely filled outershell and full energy level 3) in NACl.

So, a duplet or Octet structure is the answer

8 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$