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elena55 [62]
3 years ago
13

9. According to an "alternative theory", H2O is

Chemistry
1 answer:
Jlenok [28]3 years ago
6 0

Answer:(4) ----accepts  a proton

Explanation:

H2O water can produce both hydrogen and   hydroxide ions

H2O --> H+ + OH-

According to the Bronsted-Lowry theory, it can be a proton donor and a proton acceptor.this means that It can donate a hydrogen ion to become its conjugate base, or  can accept a hydrogen ion to form its conjugate acid,

When , a water molecule, H2O accepts a proton it will act as a Brønsted-Lowry base especially when dissolved in a strong acidic medium. for eg

HCl + H2O(l) → H3O+(aq) + Cl−(aq)

Here, Hydrochloric acid is a strong acid and ionizes completely in  water, since it is more acidic than water, the water will act as a base.

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Consider the SCl2 molecule. (a) What is the electron config- uration of an isolated S atom? (b) What is the electron con- figura
iragen [17]

Answer:

(a) 1s² 2s² 2p⁶ 3s² 3p⁴

(b) 1s² 2s² 2p⁶ 3s² 3p⁵

(c) sp³

(d) No valence orbital remains unhybridized.

Explanation:

<em>Consider the SCl₂ molecule. </em>

<em>(a) What is the electron configuration of an isolated S atom? </em>

S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.

<em>(b) What is the electron configuration of an isolated Cl atom? </em>

Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.

<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>

SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.

<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>

No valence orbital remains unhybridized.

7 0
3 years ago
A car is traveling 83.3 kilometers per hour. What is its speed in miles per minute? One kilometer = 0.62 miles.
Leya [2.2K]
Hi. 2fouls! :)

First convert the number from kilometer to miles:
83.3*0.62
=51.646 miles

Now to find the miles per minutes,divide the miles per hours by 60 (the amount of minutes in an hour.)

51.646/60
=0.860766667

The car is travelling at a speed of 0.860766667 miles/minute.

Hope this helps.
-Benjamin

3 0
3 years ago
Read 2 more answers
50 POINTS*** Which of the following is not a correct chemical equation for a double displacement reaction?
hammer [34]
I mostly believe in between D and B beacuse K3po4 and caco3 is not an element equation

4 0
3 years ago
Read 2 more answers
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
3 years ago
If star a appears to move back and forth by a greater amount than star b, which star is closer to you?
Sveta_85 [38]

If Star A appears to move back and forth by a greater amount than Star B, which star do you think is actually closer to you? Star A. If the parallax angle for a star is 1 arcsecond, what is the distance from the Sun to that star

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