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2 years ago

Describe how two atoms come together to form a single covalent bond.

1 answer:

8 0

When an atom donates electrons and other gains electrons, they form ions and the bond is called the ionic bond. When the atoms share electrons between them, they form a molecule, and the bond is called the covalent bond.

The atoms can share one, two or three pairs of electrons in the same bond. If one pair is shared it's called a single bond or a covalent simple; if two pairs are shared it's called a double bond or a covalent double, and if three pairs are shared it's called a triple bond or a covalent triple.

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Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.

N76 [4]

Answer:

1 . $Al_2O_3$

2. $TiO_2$

Explanation:

The more stable the ionic compound, the more is it lattice energy.

The more the charge on the cation and the anion, the greater is the lattice energy.
The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide ( $Sc_2O_3$ ) is an oxide in which $Sc^{3+}$ behaves as cation and $O^{2-}$ behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . $Al_2O_3$

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation $Al^{3+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

2. $TiO_2$

This is because the charge on the cation $Ti^{4+}$ is greater than that of $Sc^{3+}$ and also the size of the cation $Ti^{4+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

3 0

2 years ago

If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?

Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

Volume of Hexane dispensed = 40ml

Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

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7 0

1 year ago

Which periodic trend quantifies the amount of energy required to remove an electron from a neutral, gaseous atom?.

igor_vitrenko [27]

Ionization energy refers to the amount of energy needed to remove an electron from an atom. Ionization energy decreases as we go down a group. Ionization energy increases from left to right across the periodic table.

<h3>What is ionization energy?</h3>

Ionization is the process by which ions are formed by the gain or loss of an electron from an atom or molecule.

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.

Smaller is the size of an atom more will be the force of attraction between its protons and electrons. Hence, more amount of energy is required to remove an electron.

Thus, we can conclude that the energy required to remove an electron from a gaseous atom is called ionization energy.

Learn more about the ionization energy here:

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7 0

1 year ago

An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate th

Anettt [7]

Answer:

The method is accurate in the calculation of the $Cu^+2$

Explanation:

As a first step we have to calculate the <u>average concentration </u>of $Cu^+2$ find it by the method.

$\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm$

Then we have to find the<u> standard deviation:</u>

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359$

For the confidence interval we have to use the formula:

μ=Average± $\frac{t*s}{\sqrt{n} }$

Where:

t=t student constant with 95 % of confidence and 5 data=2.78

μ= $0.79$ ± $\frac{2.78*0.0359}{\sqrt{5} }$

upper limit: 0.84

lower limit: 0.75

If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.