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2 years ago

Calculate the pH of the buffer

Chemistry

1 answer:

77julia77 [94]2 years ago

5 0

Answer:

Less than 7 so it is going to be acidic.

Explanation:

A buffer solution is one that resists changes in pH when small quantities of an acid or an alkali are added to it. An acidic buffer solution is simply one which has a pH of less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt.

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What are the three types of symbiotic relationships?

krok68 [10]

Answer: mutualism, commensalism, and parasitism

Explanation: with mutualism, both partners benefit. With commensalism, only one species benefits while the other is neither helped nor harmed. With parasitism, one organism (the parasite) gains benefits, while the other (host) suffers.

6 0

2 years ago

Read 2 more answers

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

5. How much heat (in calories) is absorbed by a reaction when

Wewaii [24]

Answer:

1.2 × 10⁴ cal

Explanation:

Given data

Mass of water (m): 300 g

Initial temperature: 80 °C

Final temperature: 40°C

We can calculate the heat released by the water ( $Q_w$ ) when it cools using the following expression.

$Q_w = c \times m \times (T_f - T_i)$

where

c is the specific heat capacity of water (1 cal/g.°C)

$Q_w = \frac{1cal}{g.\°C} \times 300g \times (40\°C - 80\°C) = -1.2 \times 10^{4} cal$

According to the law of conservation of energy, the sum of the heat released by the water ( $Q_w$ ) and the heat absorbed by the reaction ( $Q_r$ ) is zero.

$Q_w + Q_r = 0\\Q_r = -Q_w = 1.2 \times 10^{4} cal$

7 0

2 years ago

Calculate the Molarity when a 6.11 mL solution of 0.1 H2SO4 is diluted with 105.12 mL of water

barxatty [35]

Molarity after dilution : 0.0058 M

<h3>Further explanation </h3>

The number of moles before and after dilution is the same

The dilution formula

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution

V₁ = volume of the solution before dilution

M₂ = Molarity of the solution after dilution

V₂ = Molarity volume of the solution after dilution

M₁=0.1 M

V₁=6.11

V₂=105.12

$\tt M_2=\dfrac{M_1.V_1}{V_2}=\dfrac{0.1\times 6.11}{105.12}=0.0058~M$

5 0

3 years ago

Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5

FinnZ [79.3K]

The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)

3 0

3 years ago