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sergey [27]
2 years ago
4

You need to prepare 100.0 ml of a ph 4.00 buffer solution using 0.100 m benzoic acid (pka = 4.20) and 0.120 m sodium benzoate. h

ow many milliliters of each solution should be mixed to prepare this buffer?
Chemistry
1 answer:
ololo11 [35]2 years ago
5 0
First, we have to use Henderson-Hasselblach equation to get the weak base/weak acid ratio:

PH = Pka + ㏒[A]/[HA]
when PH = 4

Pka = 4.2 

A  is the benzoate

HA is the benzoic acid

so,
4 = 4.2 + ㏒[A]/[HA]

∴[A]/[HA] = 0.631

when [A]/[HA] = 0.12 m * VA / 0.1 m * VHA

∴0.631 = 0.12m* VA / 0.1m * VHA

now, when we have the total volume of the solution is 100 mL

∴VA = 100 mL - VHA  ( we can substitute with 100 - VHA instead of VA)

∴0.631 = 0.12 * (100-VHA) / 0.1 * VHA

we can assume that VHA = X 

so,
0.631 = 0.12 * (100-X) / 0.1 * X  by  solving this equation for X

∴X = 65.5 

∴VHA = 65.5 mL

so, VA = 100mL - 65.5 ML 
           = 34.5 mL
 
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The dissociation of calcium carbonate has an equilibrium constant of Kp = 1.16 at 1073 K . CaCO3(s) ⇄ CaO(s) + CO2(g) If you pla
tino4ka555 [31]

Answer:

<h3>Pressure of CO_2 in the container=1.6 atm</h3>

Explanation:

First balance the chemical equation:

CaCO_3(s) ⇄  CaO(s) + CO_2(g)

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

K_p=[CO_2]

[CO_2]=p

K_p=p

p=K_p = 1.16atm

Pressure of CO_2 in the container=1.6 atm

8 0
2 years ago
If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.
Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water =  Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0
2 years ago
A sample of bismuth weighing 0.687 g was converted to bismuth chloride by reacting it first with HNO3, and then with HCI, follow
alisha [4.7K]

Answer:

The empirical formula is BiCl3

% Bi = 66.27 %

Explanation:

Step 1: Data given

Mass of bismuth = 0.687 grams

Mass of bismuth chloride produced = 1.032 grams

Molar mass of bismuth = 208.98 g/mol

Molar mass of bismuth chloride = 315.33 g/mol

Step 2: The balanced equation

Step 3: Calculate moles of Bi

Moles Bi = mass Bi / molar mass Bi

Moles Bi = 0.687 grams / 208.98 g/mol

Moles Bi =  0.00329 moles

Step 4: Calculate moles of Cl

Mass of Cl = 1.032 - 0.687  = 0.345 moles

Moles Cl = 0.345 moles / 35.45 g/mol

Moles Cl = 0.00973 moles Cl

Step 5: Calculate mol ratio

We divide by the smaller number of moles:

Bi: 0.00329 / 0.00329 = 1

Cl: 0.0097The empirical formula is BiCl33/0.00329 = 3

Step 6: Calculate molar mass of BiCl3

Molar mass = 208.98 + 3*35.45 = 315.33 g/mol

Step 7: Calculate percent of Bi

% Bi = (208.98 / 315.33) * 100%

% Bi = 66.27 %

3 0
2 years ago
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
2 years ago
If an atom has sp3d2 hybridization in a molecule:
never [62]

Answer:

a. the maximum number of σ bonds that the atom can form is 4

b. the maximum number of p-p bonds that the atom can form is 2

Explanation:

Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated  is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.

The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.

This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.

And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.

4 0
2 years ago
Read 2 more answers
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