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2 years ago

Describe how the pH of a solution can be calculated

Chemistry

1 answer:

hjlf2 years ago

6 0

Answer:

The pH of an aqueous solution is the measure of how acidic or basic it is. The pH of an aqueous solution can be determined and calculated by using the concentration of hydronium ion concentration in the solution.

Explanation:

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4. Consider the following half-reactions: MnO4–(aq) + 8H+(aq) + 5e– → Mn+2(aq) + 4H2O(l) NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O

jeyben [28]

Explanation:

The chemical reaction given in the question is as follows -

MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)

NO₃⁻ (aq) + 4H⁺ (aq) + 3e⁻ → NO (g) + 2H₂O (l)

As we know , the value for reduction potential are -

Mn²⁺ = + 1.51 V

NO₃⁻ = +0.96 V

From , the data given above , the value of the reduction potential of NO₃⁻ is less than the reduction potential of Mn²⁺ .

Hence ,

NO₃⁻ can not oxidize Mn²⁺ .

5 0

2 years ago

Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.

JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

H: 1.008;
Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

Hydrogen gas $\rm H_2$ , and
Magnesium chloride, which is a salt.

The chemical equation will be something like

$\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}]$ ,

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of $\rm H_2$ that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be $\rm MgCl_2$ .

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq)$ .

Balance the equation. $\rm MgCl_2$ contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

$\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq)$ .

The number of $\rm Mg$ and $\rm Cl$ atoms shall be the same on both sides. Therefore

$\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq)$ .

The number of $\rm H$ atoms shall also conserve. Hence the equation:

$\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq)$ .

How many moles of HCl are available?

$M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}$ .

$\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol$ .

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of $\rm HCl$ is 2 while the coefficient in front of $\rm H_2$ is 1. In other words, it will take two moles of $\rm HCl$ to produce one mole of $\rm H_2$ . $\rm 4.52576\;mol$ of $\rm HCl$ will produce only one half as much $\rm H_2$ .

Alternatively, consider the ratio between the coefficient in front of $\rm H_2$ and $\rm HCl$ is:

$\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}$ .

$\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol$ .

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as $\rm 1.00\times 10^{5}\;Pa$ , that volume will be 22.7 liters.

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L$ , or

$V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L$ .

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

5 0

3 years ago

Given the following unbalanced equation:

antiseptic1488 [7]

<h3>Answer:</h3>

11.84 mol CoF₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 11.84 \ mol \ CoF_2$

7 0

3 years ago

How to tell if a compound is ionic or molecular?

zhenek [66]

<span>When naming compounds, the first thing you need to do is decide if the compound is ionic or molecular. *Ionic compounds will contain both metals and non-metals, or at least one polyatomic ion. *Acids will always include the (aq) symbol beside the formula, and the name will include the word acid.</span>

5 0

3 years ago

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

kompoz [17]

Answer: 0.8541 grams of HCl will be required.

Explanation: Moles can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Al(OH)_3$ = 0.610 g

Molar mass of $Al(OH)_3$ = 78 g/mol

$\text{Number of moles}=\frac{0.610g}{78g/mol}$

Number of moles of $Al(OH)_3$ = 0.0078 moles

The reaction between $Al(OH)_3$ and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$