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grigory [225]
2 years ago
9

A buffer is prepared by adding 100 mL of 0.50 M sodium hydroxide to 100 mL of 0.75 M propanoic acid. Is this a buffer solution,

and if so, what is its pH?
Chemistry
1 answer:
Yanka [14]2 years ago
7 0
The answer: is yes, It is a buffer solution.

first, we need to get moles of sodium hydroxide and propanoic acid:

moles NaOH = molarity * volume 
                       
                       = 0.5M * 0.1 L = 0.05 moles

moles propanoic acid = molarity * volume

                                     = 0.75 M * 0.1 L = 0.075 moles

[NaOH] at equilibrium = 0.05 m

[propanoic acid ] at equilibrium = 0.075 - 0.05 = 0.025 m

when Pka for propanoic acid (given) = 4.89 

so by substitution:

∴PH = Pka + ㏒[NaOH]/[propanoic acid ]

∴ PH = 4.89 + ㏒ 0.05 / 0.025

         = 5.19
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You are riding a bicycle home from school. You average 5 meters per second and you are headed south. What motion can you determi
aleksklad [387]

Answer : The correct option is, (D) Velocity

Explanation :

Velocity : It measures the speed of an object in a particular direction. It is a vector quantity. It depends on the magnitude and the direction. The S.I unit of velocity is, m/s

Distance : It measures the length between the two objects. The S.I unit of distance is, meter.

Speed : It is defined as the distance traveled per unit of time. It does not have a direction. The S.I unit of speed is, m/s

Acceleration : It is defined as the rate of change of velocity of an object with respect to time. The S.I unit of acceleration is, m/s^2

Hence, the motion we can determine from the given data is, velocity.

4 0
1 year ago
A mixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distilled through a simple distillation apparatus; ass
Dvinal [7]

Answer:

61 mole % propylcyclohexane and 39 mole % propylbenzene

Explanation:

For convenience, let’s call propylcyclohexane <em>Component 1 </em>and propylbenzene <em>Component 2</em>.

According to <em>Raoult’s Law</em>,  

p_{1} = \chi_{1}p_{1}^{\circ} and

p_{2} = \chi_{2}p_{2}^{\circ}

where

<em>p</em>₁ and <em>p</em>₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole \fractions of the components

<em>p</em>₁° and <em>p</em>₂° are the vapour pressures of the pure components.

So,  

p_{1} = 0.60 \times \text{769 torr} = \text{ 461 torr}

p_{2} = 0.40 \times \text{725 torr} = \text{ 290 torr}

p_{\text{tot}} = <em>p</em>₁ + <em>p</em>₂= 461 torr + 290 torr = 751 torr

∴ In the vapour

\chi_{1} = \frac{p_{1}}{p_{\text{tot}} } = \frac{\text{461 torr} }{\text{751 torr}} = 0.61

χ₂ = 1 – χ₁ = 1 - 0.61 = 0.39

8 0
1 year ago
f 23.6 mL of 0.200 M NaOH is required to neutralize 10.00 mL of a H3PO4 solution , what is the concentration of the phosphoric a
Eduardwww [97]

Answer:

Explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of  0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

<h3>The concentration of unknown phosphoric acid is  0.157M</h3>
7 0
1 year ago
Read 2 more answers
En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
MAXImum [283]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

Con ello, los kilogramos de metano que cuestan 0,45 €:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
2 years ago
Barium sulfate, BaSO4, is a white crystalline solid that is insoluble in water. It is used by doctors to diagnose problems with
pshichka [43]

Answer:2

Explanation:

Ba(OH)2 contains two oxygen atoms

BaSO4 contains four oxygen atoms.

This means that barium sulphate contains two more oxygen atoms than barium hydroxide in its formula. This is clearly seen from the two formulae shown above.

5 0
1 year ago
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