All categories

2 years ago

Determine the [OH⁻] concentration in a 0.344 M Ca(OH)₂ solution.

Chemistry

1 answer:

slavikrds [6]2 years ago

8 0

Answer:

$0.688M$

Explanation:

Hello,

In this case, it is widely acknowledged that strong bases usually correspond to those formed with metals in groups IA and IIA which have relatively high activity and reactivity, therefore, when they are dissolved in water the following dissociation reaction occurs (for calcium hydroxide):

$Ca(OH)_2\rightarrow Ca^{2+}+2OH^-$

In such a way, for the same volume, we can compute the concentration of hydroxyl ions by simple stoichiometry (1:2 molar ratio):

$0.344\frac{molCa(OH)_2}{L}*\frac{2molOH^-}{1molCa(OH)_2} \\\\0.688\frac{mol OH^-}{L}$

Or simply:

$0.688M$

Regards.

You might be interested in

In the first order reaction A → products, the initial concentration of A is 0.1 108M, and 44 s later, 0.0554M. What is the initi

lilavasa [31]

Answer : The initial rate of the reaction is, $1.739\times 10^{-3}s^{-1}$

Explanation :

First we have to calculate the rate constant of the reaction.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken for the process = 44 s

$[A_o]$ = initial amount or concentration of the reactant = 0.1108 M

$[A]$ = amount or concentration left time 44 s = 0.0554 M

Now put all the given values in above equation, we get:

$k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}$

$k=0.0157$

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0, $[A]=[A_o]$

Initial rate = 0.0157 × 0.1108 = $1.739\times 10^{-3}s^{-1}$

Therefore, the initial rate of the reaction is, $1.739\times 10^{-3}s^{-1}$

8 0

3 years ago

An acid can be defined as

ryzh [129]

An acid can be defined as a proton donor.

3 0

3 years ago

Read 2 more answers

The radius of an indium atom is 0.163 nm. If indium crystallizes in a face-centered unit cell, what is the length of an edge of

adoni [48]

<u>Answer:</u> The edge length of the unit cell is 0.461 nm

<u>Explanation:</u>

We are given:

Atomic radius of iridium = 0.163 nm

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

$a=2\sqrt{2}R$

Putting values in above equation, we get:

$a=2\sqrt{2}\times 0.163=0.461nm$

Hence, the edge length of the unit cell is 0.461 nm

8 0

2 years ago

A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0

3 years ago

How many grams are contained in a 0.183 mol sample of ammonium phosphate?

soldi70 [24.7K]

Molar mass of ( NH₄)₃PO₄ = 14.01×3 + 1.01×12 + 30.97 + 16.00×4 = 149.12 g/mol. Mass of 0.183 mol ...

7 0

2 years ago