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2 years ago

How many moles of HCl are required to neutralize 20.0 mL of 1.5 M KOH?

1 answer:

8 0

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What is the molarity of 2.3 mol of Kl dissolved in 0.5 L of water

Nastasia [14]

Answer:

$4.6\,\,moL\,\,L^{-1}$

Explanation:

Molarity refers to a measure of concentration.

Molarity = moles of solute/Litres of solution

Molarity refers to number of moles of solute present in this solution.

In order to find a solution's molarity, use value for the number of moles of solute and the total volume of the solution expressed in liters

As molarity of 2.3 mol of Kl is dissolved in 0.5 L of water,

Molarity = $\frac{2.3}{0.5} =4.6\,\,moL\,\,L^{-1}$

4 0

3 years ago

What is true of intercellular signals that do not go through direct connections between cells, such as plasmodesmata?

Ronch [10]

I believe that what happens here is that when those cells carrying those intercellular signals do no pass through the connections, actually the signal is passed through or are transferred to adjacent cells for signal transmission, hence the answer is:

6 0

3 years ago

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A particular voltaic cell operates with the reaction: Zn(s)+Cl2(g)--->Zn^2+(aq)+2Cl^-(aq) giving a cell potential of .853 V.

nadya68 [22]

Answer:

The amount of energy liberated will be 49.38 J.

Explanation:

The amount of energy liberated (gibbs free energy) can be calculated using the following equation:

ΔG° = -nFε

n: amount of moles of electrons transfered

F: Faraday's constant

ε: cell potential

20.0 g of Zn is equal to 0.30 mol.

Two electrons are transfered during the reaction.

Therefore, n = 2x0.30 ∴ n = 0.60

ΔG° = - 0.60 x 96.485 x 0.853

ΔG° = 49.38 J

3 0

3 years ago

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

IRINA_888 [86]

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:

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

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

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.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

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,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

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

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

3 years ago

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Calculate the number of grams in magnesium present in 25 g of magnesium phosphate

olga55 [171]

you can google it and it pops up right away

4 0

3 years ago

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