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3 years ago

How many molecules are in 145.5 grams of Be(OH)2

Chemistry

1 answer:

Zanzabum3 years ago

3 0

Answer:

2.04 x 10²⁴ molecules

Explanation:

Given parameters:

Mass of Be(OH)₂ = 145.5g

To calculate the number of molecules in this mass of Be(OH)₂ we follow the following steps:

>> Calculate the number of moles first using the formula below:

Number of moles = mass/molarmass

Since we have been given the mass, let us derive the molar mass of Be(OH)₂

Atomic mass of Be = 9g

O = 16g

H = 1g

Molar Mass = 9 + 2(16 + 1)

= 9 + 34

= 43g/mol

Number of moles = 145.5/43 = 3.38mol

>>> We know that a mole is the amount of substance that contains Avogadro’s number of particles. The particles can be atoms, molecules, particles etc. Therefore we use the expression below to determine the number of molecules in 3.38mol of Be(OH)₂:

Number of

molecules= number of moles x 6.02 x 10²³

Number of molecules= 3.38 x 6.02 x 10²³

= 20.37 x 10²³ molecules

= 2.04 x 10²⁴ molecules

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Vedmedyk [2.9K]

Answer:

$pH=13$

Explanation:

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In this case, given the acid, we can suppose a simple dissociation as:

$HA\rightleftharpoons H^+ + A^-$

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$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction's extent we can write:

$1x10^{-20}=\frac{x*x}{0.1M-x}$

But we prefer to compute the Kb due to its exceptional weakness:

$Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}$

Next, the acid dissociation in the presence of the base we have:

$Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}$

Whose solution is $x=0.0999M$ which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

$pOH=-log([OH^-])=-log(0.0999)=1$

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$pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13$

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3 years ago

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You are given 3 unknown solutions with pH value as 6, 8 & 9.5 respectively. Which solution will contain maximum OH ion

Grace [21]

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3 years ago

Making solutions is an extremely important component to real-life chemistry. If you make 3.00 L of a solution using 90.0 g of so

kumpel [21]

Answer:

Final concentration of NaOH = 0.75 M

Explanation:

For $NaOH$ :-

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$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{90.0\ g}{39.997\ g/mol}$

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The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Where, Volume must be in Liter.

It is denoted by M.

Given, Volume = 3.00 L

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$Molarity=\frac{2.2502\ mol}{3.00\ L}=0.75\ M$

<u>Final concentration of NaOH = 0.75 M</u>

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4 years ago

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castortr0y [4]

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