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3 years ago

Balancing Chemical Equations

1 answer:

Citrus2011 [14]3 years ago

6 0

Start by assigning the most complicated species a coefficient of one. That species should contain the greatest number of elements, e.g. $\text{HCl}$ .

$\text{Zn} + \text{HCl} \to 1\; \text{ZnCl}_2 + \text{H}_2$ (not balanced)

Assign coefficients to the rest of the species based on the conservation of atoms. For instance, the left hand side of the equation now contains one atom of hydrogen H and one atoms of chlorine Cl. The left hand side shall have an identical configuration. Both zinc chloride $\text{ZnCl}_2$ and hydrogen $\text{H}_2$ should therefore have a coefficient of 1/2. (Don't panic about the fractions. They are to be eliminated in a few more steps.)

$\text{Zn} + 1\; \text{HCl} \to 1/2 \; \text{ZnCl}_2 + 1/2 \; \text{H}_2$ (not balanced)

The coefficient 1/2 in front of zinc chloride indicates the presence of 1/2 zinc atom in the right hand side of the equation. Zinc on the left hand side of the equation should accordingly have a coefficient of 1/2.

$1/2 \; \text{Zn} + 1\; \text{HCl} \to 1/2 \; \text{ZnCl}_2 + 1/2 \; \text{H}_2$

Increasing coefficients on both sides of the equation by a factor of two to eliminate all fractions. Hence the balanced equation.

$1 \; \text{Zn} + 2 \; \text{HCl} \to 1 \; \text{ZnCl}_2 + 1 \; \text{H}_2$ (balanced)

The same set of operations should work for the second equation.

$4 \; \text{Fe} + 3\; \text{O}_2 \to 2\; \text{Fe}_2\text{O}_3$ (balanced)

Note that the third equation does not accurately represent the catalytic decomposition of hydrogen peroxide $\text{H}_2\text{O}_2$ . The balanced equation should be:

$2\; \text{H}_2\text{O}_2 \to 2\; \text{H}_2\text{O} + \text{O}_2$ (balanced)

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A 50.0 50.0 mL solution of 0.127 0.127 M KOH KOH is titrated with 0.254 0.254 M HCl HCl. Calculate the pH of the solution after

Brilliant_brown [7]

Answer:

Explanation:

THE CORECT QUESTION

A 50.0 mL solution of 0.127 M KOH is titrated with 0.254 M HCl. Calculate the pH of the solution after the addition of each of the given amounts of HCl.

SOLUTION

Get the concentration of the HCl first using titration formula

CA X V A / CB VB = NA/ NB

Equation of reation; KOH + HCl => KCl + H2O

CA = 0.254 M

CB = 0.127

VA = 1/0.254 = 3.937

CA (after the addition) = 0.127 x 50 / 3.937

= 1.612 M

But pH = - Log[hydrogen ion]

= -log 1.612

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2 years ago

A mass spectrometer has determined the mass and abundances of all isotopes of an unknown element. The first isotope has a mass o

shepuryov [24]

Answer:

I believe it is Potassium (K)

Explanation:

I did the math on a calculator and it was the closest atomic mass to potassium.

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3 years ago

A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

seropon [69]

Answer: The p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, $M_1=(2\times 0.0031)=0.0062M$

Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, $M_2=(2\times 0.0042)=0.0084M$

To calculate the concentration of nitrate ions in the solution, we use the equation:

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

Calculating the p-function of zinc ions and nitrate ions in the solution:

For zinc ions:

$\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]$

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

For nitrate ions:

$\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]$

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

Hence, the p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

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3 years ago

What is responsible for the larger size of an anion in comparison with the atom from which it is form?

GarryVolchara [31]

Answer:

electron-electron repulsion

Explanation:

When electrons add into valence shell of neutral elements, the element assumes a negative oxidation state. With this, the number of electrons having (-) charges will be larger than the number of protons having positive (+) charges. As a result, the extra electrons repel one another (i.e., like charges repel) and a larger radius is the result.

In contrast, when cations are formed, electrons are removed from the valence level (oxidation) producing an element having a greater number of protons than electrons. The larger number of protons will function to attract the electron cloud with a greater force that results in a contraction of atomic radius and a smaller spherical volume than the neutral unionized element.

To visualize, see attached chart that shows atomic and ionic radii before and after ionization of the elements.

Download pdf

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Question 1 Which of the following is a false

Bogdan [553]

Answer:

1.A 2.C 3.B

4. d

Explanation:

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