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3 years ago

What is molarity???????

Chemistry

1 answer:

velikii [3]3 years ago

4 0

Molarity is a term that applies to the amount of Moles of solute, per each litre of the solution.

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What is conjugate solution

mars1129 [50]

Answer:

Explanation: A mixture of two partially miscible liquids

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3 years ago

Which part of the fuel cell does A represent?

Bad White [126]

I think the anwer is electrolyte :)... i had it on a test a couple days ago.

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3 years ago

Read 2 more answers

What would you need to do to calculate the molality of 10 g of NaCl in 2 kg of

jasenka [17]

Answer:

O B. Convert the 10 g of NaCl to moles of NaCl.

Explanation:

The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.

(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl

58.44 is the molar mass of NaCl

m=0.1711 mol NaCl/2 kg H2O

m=0.085557837

7 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

3 years ago

What affects the rates of reactions? describe each. <br> a. <br> b.?

Anvisha [2.4K]

Rates can be calculated with Arrhenius equation k = Axe^(-Ea/RT)
a. temperature affects the rate - imagine you are making coffee, so coffee crystals are boiled faster on higher temperature. Simplified but makes sense.
b. Ea is activation energy. Imagine, while preparing coffee, some of ingredients change to a different one, so there is a A -> B reaction (simplified). Now, Ea is energy barrier that stands on the arrow of this reaction, preventing A to transform to B. If Ea is small, reaction will go easy (not fast!), if Ea is large –reaction will not happen so easy (you ll have to use catalyst for example)

4 0

3 years ago