Which element is found in all the acids described in this section?

How is electronegativity related to covalent bonding?

Zolol [24]

Answer:

Atoms must have similar electronegativities in order to share electrons in a covalent bond.

Explanation:

Covalent bonding is one of the bondings that occurs between the atoms of elements. It is the bonding in which atoms share their valence electrons with one another. However, the ELECTRONEGATIVITY, which is the ability of an atom to be attracted to electrons play a major role in the formation of covalent bonds.

When atoms of different electronegativities combine, the more electronegative atom pulls more electrons towards itself, hence, an IONIC bond is formed. However, when the electronegativities of the atoms are similar, the sharing of their electrons becomes stronger. Hence, ATOMS MUST HAVE SIMILAR ELECTRONEGATIVITIES in order to share electrons in a covalent bond.

5 0

3 years ago

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3 & 4 Help pleaseeeee

kvasek [131]

3 is C i believe, and as for 4 minerals are solid, natural.. that's about all I can say...

4 0

3 years ago

How many atoms are there in 8.88 g Si?

Mariana [72]

Answer:

$\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}$

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

Si: 28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

$\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

$8.88 \ g \ Si *\frac { 28.085 \ g \ Si}{1 \ mol \ Si}$

Flip the fraction so the units of grams of silicon cancel.

$8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \ Si}$

$8.88 *\frac{1 \ mol \ Si} { 28.085 }$

$\frac {8.88} { 28.085 } \ mol \ Si$

$0.316183015845 \ mol \ Si$

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

$\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

Multiply by the number of moles we calculated.

$0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}$

The units of moles of silicon cancel.

$0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}$

$0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}$

$1.90405412 \times 10^{23} \ atoms \ Si$

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

$1.90 \times 10^{23} \ atoms \ Si$

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0

3 years ago

Calculate the standard free-energy change at 25 ∘C for the following reaction:

lianna [129]

Answer:

Standard free-energy change at $25^{0}\textrm{C}$ is $-3.80\times 10^{2}kJ/mol$

Explanation:

Oxidation: $Mg(s)-2e^{-}\rightarrow Mg^{2+}(aq.)$

Reduction: $Fe^{2+}(aq.)+2e^{-}\rightarrow Fe(s)$

--------------------------------------------------------------------------------------

Overall: $Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)$

Standard cell potential, $E_{cell}^{0}=E_{Fe^{2+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}$

So, $E_{cell}^{0}=(-0.41V)-(-2.38V)=1.97V$

We know, standard free energy change at $25^{0}\textrm{C}$ ( $\Delta G^{0}$ ): $\Delta G^{0}=-nFE_{cell}^{0}$

where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol

Here n = 2

So, $\Delta G^{0}=-(2)\times (96500C/mol)\times (1.97V)=-380210J/mol=-380.21kJ/mol=-3.80\times 10^{2}kJ/mol$

8 0

3 years ago

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If the caterpillar keeps moving at the same speed, how long will it take to travel 6cm?

Solnce55 [7]

It will take 15 s to travel 6 cm

<h3>Further explanation</h3>

Given

distance versus time graph

Required

time travel

Solution

Caterpillar motion is a straight motion with a constant speed, so that the graph between distance and time forms a diagonal line

If we look at the graph, we can determine the time taken when the distance reaches 6 cm (y axis) by drawing a line to the diagonal line and cutting the x-axis as time, and we get 15 s

Or we can also use the formula for motion at constant speed:

d = v x t

With v at point 2,5 of 2/5 m / s, so the time taken:

$\tt t=\dfrac{d}{v}=\dfrac{6}{2/5}=15~s$

4 0

3 years ago