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Soloha48 [4]
3 years ago
10

A 0.005 M potassium hydroxide solution is an example of what type of solution? *

Chemistry
1 answer:
poizon [28]3 years ago
7 0

Answer:

Basic solution

Explanation:

A solution containing 0.005 M potassium hydroxide has a pH of;

pOH= -log[0.005]

pOH= 2.3

But pH+pOH=14

Therefore;

pH= 14-pOH

Hence;

pH= 14-2.3= 11.7

A solution having a pH of 11.3 is clearly a basic solution because this pH indicates a substance that is basic in nature. Hence the answer given.

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a radiator is filled with a mixture of 3.25kg ethylene glycol (C2H6O2) in 7.75kg of water. calculate the molality of this questi
Vladimir [108]
3.25 kg in g = 3.25 * 1000 = 3250 g

Molar mass C₂H₆O₂ =  62.0 g/mol

Mass solvent = 7.75 kg

Number of moles:

n = mass solute / molar mass

n = 3250  / 62.0 

n = 52.419 moles

Molality = moles of solute / kilograms of solvent

M = 52.419 / 7.75

M = 6.7637 mol/kg

hope this helps!
5 0
3 years ago
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
______ is the use of enzymes is necessary to increase the activation energy requirements of a chemical reaction.
Masja [62]

Answer:

Catalysts

Explanation:

Catalysts lower the aviation energy.

6 0
3 years ago
A scientist has two samples of gas: The first sample contains one mole of argon atoms and has a mass of 39.948 g; the second sam
V125BC [204]

There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.

Explanation:

39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.

4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.

We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.

So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.

Learn more about:

Avogadro's number

brainly.com/question/14148121

brainly.com/question/1445383

brainly.com/question/1528951

#learnwithBrainly

4 0
4 years ago
Identify the right
Basile [38]

.answer:

so the first one is analogous

the second is homologous

and the third is analogous

explanation:

analogous. this means they share a similar function (flight) but do not have the same embryonic origin.

6 0
3 years ago
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