Draw the structure of the following atoms<br> (1) 19/9 F<br> (2)28/14 SI

Hat volume of a 0.540 M NaOH solution contains 12.5 g NaOH

amid [387]

Answer:

At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.

Explanation:

Given data:

At volume = ?

Mass of NaOH = 12.5 g

Molarity of solution = 0.540 M

Solution:

First of all we will calculate the number of moles of sodium hydroxide.

Number of moles = mass/molar mass

Number of moles = 12.5 g / 40 g/mol

Number of moles = 0.3125 mol

Volume of NaOH:

Molarity = number of moles / volume in L

Now we will put the values.

0.540 M = 0.3125 mol / volume in L

volume in L = 0.3125 mol / 0.540 mol/L

volume in L = 0.58 L

5 0

3 years ago

How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?

Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

$pH= -log[H]\\pH= -log (\frac{kw}{[OH]})$

$8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}$

$[OH]=1.69E-6$

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

$Cu(OH)_2 -> Cu^{2+} + 2OH^-$

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

$Kps= [Cu^{2+}] [OH]^2$

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

$Kps= s*(2s+1.69E-6)^2$

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is $s=4.96E-8 mol/L$

Now, let us calculate the moles in 1 L:

$moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles$

7 0

4 years ago

PLEASE HELP SOMEONE PLEASEEEE

crimeas [40]

Answer:

the answer is b

hope this helps you!!

Have a lovely day!

-millie :D

4 0

3 years ago

How much sodium chloride did you add?

MAVERICK [17]

Answer:

add 1/2 to 3/4 cup salt (sodium chloride) to the gallon ziploc bag of ice.

7 0

3 years ago

For a second order reaction, the half-life is equal to:_____.

gtnhenbr [62]

Answer:

Half life = 1 / k[Ao]

Explanation:

From:

1/ [A] = kt + 1/ [Ao]

Isolating t on its own, we have:

kt = 1 / [A] - 1 / [Ao]

t = 1 / [Ao] / k

Re-arranging we have:

t = 1 / k [Ao]

The t represents the t=half life of the second order reaction and the formula can be re-written as:

t1/2 = 1 / k [Ao]

This is so because second order reaction decreases at a much faster rate than zero and first order reactions and there slopes decreases to zero at a much faster rate.

4 0

3 years ago

Draw the structure of the following atoms (1) 19/9 F (2)28/14 SI