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3 years ago

Every conversion factor is a ratio that is equal to 1 less than 1 greater than 1

Chemistry

1 answer:

miss Akunina [59]3 years ago

4 0

equal to 1

Explanation:

All conversion factors used in any calculation is usually a ratio that is equal to 1. This is because when 1 is used to multiply or divide any number, the number stays the same.

Conversion factors are useful in making the simplification of an expression very simple. They are usually ratios of 1.
It follows that a conversion factor is a number that when multiplied or divided by itself will stay the same.

Learn more:

conversion factor: brainly.com/question/555814

#learnwithBrainly

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You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,

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Answer:

$\frac{[F^{-}]}{[HF]}$ is larger

Explanation:

$pK_{a}=-logK_{a}$ , where $K_{a}$ is the acid dissociation constant.

For a monoprotic acid e.g. HA, $K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}$ and $\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}$

So, clearly, higher the $K_{a}$ value , lower will the the $pK_{a}$

In this mixture, at equilibrium, $[H^{+}]$ will be constant.

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

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Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

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Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

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The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

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