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Reil [10]
3 years ago
7

Cómo podemos aplicar los conocimientos químicos a nuestra vida cotidiana?

Chemistry
1 answer:
lianna [129]3 years ago
7 0
Cause of imagination
저 하늘의 오렌지빛 마법이 끝이 나기 전에
'Cause of imagination 그 찰나에
Can you feel the rush? Can you feel the rush?
시간아 멈춰줘
I wanna stay, hey (Can you feel the rush? Can you feel the rush?)
두 세계의 경계선 그 틈에 너와 나를 남기고파 (Come on, let's do it)
I wanna stay, hey (Can you feel the rush? Can you feel the rush?)
개와 늑대의 그 시간엔 마법 속에 갇히고파 (Blue hour!)
You might be interested in
Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
diamong [38]

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

<u>For the steps below, refer to the ICE chart above.</u>

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}.

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

1.8*10^-^4 =\frac{x^2}{0.1-x}

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

8 0
3 years ago
NEED HELP ASAP
joja [24]

A.Shield Volcano -----



3 0
4 years ago
Place the following covalent bonds in order from least to most polar: a. H-Cl b. H-Br c. H-S d. H-C
timofeeve [1]

Answer:


The Order is as follow,

                                     C-H < S-H < H-Br < H-Cl


Explanation:

                      Polarity depends on the electronegativity difference between two atoms, greater the electronegativity difference, greater will be the polarity of bond and vice versa.


Electronegativity Difference between Hydrogen and other given elements are as follow,


1) C-H;

               E.N of Carbon     = 2.55

               E.N of Hydrogen = 2.20

                                           ------------

               Difference              0.35


2) S-H;

               E.N of Sulfur       = 2.58

               E.N of Hydrogen = 2.20

                                           ------------

               Difference               0.38


3) H-Br;

               E.N of Bromine   = 2.96

               E.N of Hydrogen = 2.20

                                          -------------

               Difference              0.76


4) H-Cl;

               E.N of Chlorine   = 3.16

               E.N of Hydrogen = 2.20

                                           -----------

               Difference               0.96


Hence it is proved that the greatest electronegativity difference is found between H and Chlorine in H-Cl, therefore it is highly polar bond and vice versa.

7 0
3 years ago
Write yes if the proper safety procedures are being followed and no if they are not. Then give a reason for your answer. Bubba c
Oksi-84 [34.3K]
<span>No. The proper safety procedure is not being enforced in this scenario. First of all, Bubba should already know the teacher's lecture on the correct and safe way to light the Bunsen burners in his/her class.</span>
4 0
3 years ago
Read 2 more answers
Calculate the solubility of Co(OH)2, in g/L, in solutions that have been buffered to the following pHs.
mr_godi [17]

Answer:

a. 14.9g/L

b. 1.49x10⁻⁶

c. 1.49x10⁷

Explanation:

You can write the buffer Ksp of Co(OH)₂ as follows:

Co(OH)₂(s) ⇄ Co²⁺ + 2OH⁻

Ksp = 1.6x10⁻¹⁵ = [Co²⁺] [OH⁻]²

To have buffered the solutions means  [OH⁻] is fixed. From the equilibrium of water we can relate  [OH⁻] with pH as follows:

[OH⁻] = 10^[14-pH]

With [OH⁻] and Ksp we can solve for [Co²⁺]. Its concentration is equal to solubility (That is the amount of Co(OH)₂ that can be dissolved).

[Co²⁺] is in mol/L. With molar mass of Co(OH)₂ -92.948g/mol-, We can obtain, in the end, its solubility in g/L.

-<em>Molar concentration of [Co²⁺] and solubility:</em>

a. [OH⁻] = 10^[14-7.00] = 1x10⁻⁷

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁷]²

[Co²⁺] = 0.16mol / L = Solubility.

In g/L = 0.16mol / L ₓ(92.948g/mol) =

14.9g/L

b. [OH⁻] = 10^[14-10.00] = 1x10⁻⁴

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻⁴]²

[Co²⁺] = 1.6x10⁻⁸mol / L = Solubility.

In g/L = 1.6x10⁻⁸mol / L ₓ(92.948g/mol) =

1.49x10⁻⁶g/L

c.[OH⁻] = 10^[14-4.00] = 1x10⁻¹⁰

[Co²⁺] = 1.6x10⁻¹⁵ / [1x10⁻¹⁰]²

[Co²⁺] = 1.6x10⁵mol / L = Solubility.

In g/L = 1.6x10⁵mol / L ₓ(92.948g/mol) =

1.49x10⁷g/L

As you can see, and as general rule, <em>all hydroxides are solubles in acids</em>.

6 0
3 years ago
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