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2 years ago

What equation is used to calculate the pH of a solution?

Chemistry

1 answer:

Galina-37 [17]2 years ago

7 0

Answer:

pH = -log[H3O+] or pH = -log[H+]

Explanation:

pH is the negative log of hydronium concentration or hydrogen ion concentration. The two are essentially interchangeable.

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A galvanic cell whose cell reaction is 2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq) has a cell potential of 0.72V. What is the maxim

sesenic [268]

Answer:

138.96kJ is the maximum electrical work

Explanation:

The maximum electrical work that can be obtained from a cell is obtained from the equation:

W = -nFE

<em>Where W is work in Joules,</em>

<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>

Zn(s) → Zn²⁺(aq) + 2e⁻

F is faraday constant = 96500Coulombs/mol

E is cell potential = 0.72V

Replacing:

W = -2mol*96500Coulombs/mol*0.72V

W = - 138960J =

<h3>138.96kJ is the maximum electrical work</h3>

<em />

5 0

2 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

2 years ago

How many moles of carbon are in the sample?

jeka94

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5 0

2 years ago

55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?

ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$ ,