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3 years ago

What do you understand by shared pair of electron

1 answer:

5 0

Answer:Lone pairs are the valence electron pair of any element which do not take part in bonding but affect the shape of molecules. Bond pairs or shared pairs are the electron pair which does both affect the geometry of molecules and take part in chemical bonding. These are form due to sharing of electrons.

Explanation:

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Consider the titration of a 20.0-mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine each quantity. a. the initial pH b. t

Oksi-84 [34.3K]

Answer:

Explanation:

Given that:

Concentration of $HC_2H_3O_2 \ (M_1)$ = 0.105 M

Volume of $HC_2H_3O_2 \ (V_1)$ = 20.0 mL

Concentration of $NaOH (M_2)$ = 0.125 M

The chemical reaction can be expressed as:

$HC_2H_3O_2_{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O_{(l)}$

Using the ICE Table to determine the equilibrium concentrations.

$HC_2 H_3 O_2 _{(aq)} + H_2O _{(l) } \to C_2 H_3O_2^- _{(aq)} + H_3O^+_{ (aq)}$

I 0.105 0 0

C -x +x +x

E 0.105 - x x x

$K_a = \dfrac{[C_2H_5O^-_2][H_3O^+]}{[HC_2H_3O_2]}$

$K_a = \dfrac{(x)(x)}{(0.105-x)}$

Recall that the ka for $HC_2H_3O_2= 1.8 \times 10^{-5}$

Then;

$1.8 \times 10^{-5} = \dfrac{(x)(x)}{(0.105 -x)}$

$1.8 \times 10^{-5} = \dfrac{x^2}{(0.105 -x)}$

By solving the above mathematical expression;

x = 0.00137 M

$H_3O^+ = x = 0.00137 \ M \\ \\ pH = - log [H_3O^+] \\ \\ pH = - log ( 0.00137 )$

pH = 2.86

Hence, the initial pH = 2.86

b) To determine the volume of the added base needed to reach the equivalence point by using the formula:

$M_1 V_1 = M_2 V_2$

$V_2= \dfrac{M_1V_1}{M_2}$

$V_2= \dfrac{0.105 \ M \times 20.0 \ mL }{0.125 \ M}$

$V_2 = 16.8 mL$

Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL

c) when pH of 5.0 mL of the base is added.

The Initial moles of $HC_2H_3O_2 =$ molarity × volume

$= 0.105 \ M \times 20.0 \times 10^{-3} \ L$

$= 2.1 \times 10^{-3}$

number of moles of 5.0 NaOH = molarity × volume

number of moles of 5.0 NaOH = $0.625 \times 10^{-3}$

After reacting with 5.0 mL NaOH, the number of moles is as follows:

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Initial moles $2.1*10^{-3}$ $0.625 * 10^{-3}$ 0 0

F(moles) $(2.1*10^{-3} - 0.625 \times 10^{-3})$ 0 $0.625 \times 10^{-3}$ $0.625 \times 10^{-3}$

The pH of the solution is then calculated as follows:

$pH = pKa + log \dfrac{[base]} {[acid]}$

Recall that:

pKa for $HC_2H_3O_2=4.74$

Then; we replace the concentration with the number of moles since the volume of acid and base are equal

∴

$pH = 4.74 + log \dfrac{0.625 \times 10^{-3}}{1.475 \times 10^{-3}}$

pH = 4.37

Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37

We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.

Therefore;

pH = pKa = 4.74

e) pH at the equivalence point.

Here, the pH of the solution is the result of the reaction in the $(C_2H_3O^-_2)$ with $H_2O$

The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)

The total volume(V) of the solution = 20.0 mL + 16.8 mL

The total volume(V) of the solution = 36.8 mL

Concentration of $(C_2H_3O^-_2)$ = moles/volume

= $\dfrac{2.1 \times 10^{-3} \ moles}{0.0368 \ L}$

= 0.0571 M

Now, using the ICE table to determine the concentration of $H_3O^+$ ;

$C_2H_5O^-_2 _{(aq)} + H_2O_{(l)} \to HC_2H_3O_2_{(aq)} + OH^-_{(aq)}$

I 0.0571 0 0

C -x +x +x

E 0.0571 - x x x

Recall that the Ka for $HC_2H_3O_2$ = $1.8 \times 10^{-5}$

$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0\times 10^{-14}}{1.8 \times 10^{-5} } \\ \\ K_b = 5.6 \times 10^{-10}$

$k_b = \dfrac{[ HC_2H_3O_2] [OH^-]}{[C_2H_3O^-_2]}$

$5.6 \times 10^{-10} = \dfrac{x *x }{0.0571 -x}$

$x = [OH^-] = 5.6 \times 10^{-6} \ M$

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{5.6 \times 10^{-6} }$

$[H_3O^+] =1.77 \times 10^{-9}$

$pH =-log [H_3O^+] \\ \\ pH =-log (1.77 \times 10^{-9}) \\ \\ \mathbf{pH = 8.75 }$

Hence, the pH of the solution at equivalence point = 8.75

f) The pH after 5.09 mL base is added beyond (E) point.

$HC_2 H_3 O_2 _{(aq)} + NaOH _{(aq)} \to NaC_2H_3O_2_{(aq)} + H_2O{ (l)}$

Before 0.0021 0.002725 0

After 0 0.000625 0.0021

$[OH^-] = \dfrac{0.000625 \ moles}{(0.02 + 0.0218 ) \ L}$

$[OH^-] = \dfrac{0.000625 \ moles}{0.0418 \ L}$

$[OH^-] = 0.0149 \ M$

From above; we can determine the concentration of $H_3O^+$ by using the following method:

$[H_3O^+] = \dfrac{1.0 \times 10^{-14} }{0.0149}$

$[H_3O^+] = 6.7 \times 10^{-13}$

$pH = - log [H_3O^+]$

$pH = -log (6.7 \times 10^{-13} )$

pH = 12.17

Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17

3 0

3 years ago

Calculate the number of C, H, and O atoms in 1.50 g of glucose, a sugar

DIA [1.3K]

Chemical formula of the glucose: C₆H₁₂O₆

We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u

atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol

2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g---------------------- x

x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles

we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)

3)We calculate the number of molecules:

Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles-------- x

x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.

4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²

number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .

number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²

Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.

4 0

3 years ago

What is The strength of a hurricane is ranked from 1 to 5 on the _________________.

UkoKoshka [18]

Richter Scale.

Hope this helps!

8 0

3 years ago

Be sure to answer all parts. Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas: 2NO(g) +

notka56 [123]

<u>Answer:</u> NO is the limiting reagent in the given reaction and 0.857 moles of $NO_2$ will be produced.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

We are given:

Moles of NO = 0.857 mol

Moles of oxygen = 0.498 mol

By stoichiometry of the reaction:

If 2 moles of NO reacts with 1 mole of oxygen gas.

So, 0.857 moles of NO will react with = $\frac{1}{2}\times 0.857=0.4285mol$ of $O_2$

As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, NO is considered as the limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

If 2 moles of NO produces 2 moles of nitrogen dioxide gas.

So, 0.857 moles of NO will produce = $\frac{2}{2}\times 0.857=0.857mol$ of $NO_2$

Hence, NO is the limiting reagent in the given reaction and 0.857 moles of $NO_2$ will be produced.

6 0

3 years ago

A 475 cm3 sample of gas at standard temperature and pressure is allowed to expand until it occupies a

Andrej [43]

The final temperature : 345 K

<h3> Further explanation </h3>

Given

475 cm³ initial volume

600 cm³ final volume

Required

The final temperature

Solution

At standard temperature and pressure , T = 273 K and 1 atm

Charles's Law :

When the gas pressure is kept constant, the gas volume is proportional to the temperature

V₁/T₁=V₂/T₂

Input the value :

T₂=(V₂T₁)/V₁

T₂=(600 x 273)/475

T₂=345 K

4 0

3 years ago

What do you understand by shared pair of electron ​

What do you understand by shared pair of electron