Answer:
Spontaneous processes are ones that occur quickly and have a low activation energy. - False -
Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly - False
Spontaneous processes can occur slowly, but always have a low activation- false
Spontaneous reactions can react slowly and can have a high activation energy - True
Spontaneous processes always react slowly and always have a high activation energy- False
Explanation:
A spontaneous reaction is reaction that proceeds on its own without us having to do a thing at all!
A spontaneous reaction may be fast or slow depending on the activation energy of the reaction. A spontaneous reaction having a high activation energy will be slow. However, if the spontaneous reaction has a low activation energy then it will be fast.
We have to note here that a spontaneous reaction proceeds without a prolonged input energy. Sometimes energy may be supplied to the reaction at the beginning for instance in the case of the combustion of hydrocarbons.
So, spontaneous processes are not necessarily fast. Some of them may have a very high activation energy such as in the rusting of iron hence they are slow.
15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
You can learn more about equilibrium constants here:
https://brainly.com/question/15118952
If 1 mol of ferric oxide reacts with 3 moles of carbon monoxide to yield 2 mols of iron and 3 mols of carbon dioxide, how much CO will be needed to completely react with 50.26 g of ferric oxide?
Answer:
26.4g
Explanation:
The balanced chemical equation as stated in this question is given as follows:
Fe2O3 + 3CO → 2Fe + 3CO2
According to this balanced equation, 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).
We need to change 50.26 g of ferric oxide to moles by using the formula;
mole = mass/molar mass
Molar mass of Fe2O3 = 56(2) + 16(3)
= 112 + 48
= 160g/mol
mole = 50.26/160
mole = 0.314mol of Fe2O3
If 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).
Hence, 0.314 mol of Fe2O3 will completely react with (0.314 × 3) mol of CO
0.314 × 3 = 0.94 mol of CO
molar mass of CO = 12 + 16 = 28g/mol
mole = mass/molar mass
mass = mole × M.M
mass = 0.94 × 28
mass = 26.4g of CO
What do we need to know to understand the formation of a chemical bond?
Answer:
A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.
Explanation:
You have to put energy into a molecule to break its chemical bonds. The amount needed is called the bond energy. After all, molecules don't spontaneously break
14. What is the oxidation number of oxygen in HSO4 -
Answer:
2
Explanation:
i did this
A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Kaof hypochlorous acid is 3.0x10^-8.
a) 10.20
b) 7.00
c) 6.48
d) 7.52
e) 14.52
Answer:
pH = 10.20
Explanation:
The HClO reacts with NaOH as follows:
HClO + NaOH → H2O + NaClO
Where HClO and NaOH react in a 1:1 reaction.
As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.
And the NaClO produced decreases its concentration in 2 because the volume is doubled.
The concentration of NaClO is: 0.150M / 2 = 0.075M
The equilibrium of NaClO is:
NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)
Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =
1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]
[NaClO] = 0.075M
As both HClO and OH- comes from the same equilibrium,
[HClO] = [OH-] = X
Where X is the reactoin coordinate
Replacing:
3.33x10⁻⁷ = [X] [X] / [0.075M]
2.5x10⁻⁸ = X²
X = 1.58x10⁻⁴M = [OH-]
pOH = -log [OH-]
pOH = 3.80
pH = 14 - pOH
pH = 10.20When naming organic compounds, there are strict rules regarding punctuation.
1. A comma is used to separate two numbers.
2. A hyphen is used to separate a number from a letter.
Rewrite the name of this compound using hyphens and commas as appropriate.
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
2,3,3-trimethylhexane
Explanation:
IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.
According to this system of nomenclature;
A comma is used to separate two numbers.
A hyphen is used to separate a number from a letter.
Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.
The density of an aqueous solution containing 25.0 percent of ethanol (C2H5OH) by mass is 0.950 g/mL. (a) Calculate the molality of this solution. m (b) Calculate its molarity. M (c) What volume of the solution would contain 0.275 mole of ethanol
Answer:
a. 7.24m
b. 5.15M
c. 53.4mL of the solution would contain this amount of ethanol.
Explanation:
Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.
Molarity, M, are the moles of solute per kg of solvent
To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:
a. Moles ethanol -Molar mass: 46.07g/mol-:
25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol
kg solvent:
100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg
Molality:
0.54265 moles ethanol / 0.075kg = 7.24m
b. Liters solution:
100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L
Molarity:
0.54265 moles ethanol / 0.1053L = 5.15M
c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =
53.4mL of the solution would contain this amount of ethanol
g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular equation, what are the products
Answer:
Where the products are H2O and Ba(NO3)2
Explanation:
A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called neutralization reaction.
The balanced reaction is:
Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2
Where the products are H2O and Ba(NO3)2
Tech A says that the PCM monitors the pre-cat and post-cat oxygen sensors to determine catalytic converter efficiency. Tech B says that a catalytic converter can be tested by graphing the oxygen sensor readings on a scan tool or lab scope and comparing them. Who is correct
Answer:
Both Tech A and Tech B.
Explanation:
Catalyst is an element used to start chemical reaction but is not used in the reaction. Catalysts material used in catalytic converter include Rhodium, Palladium and platinum. The pre cat and post cat oxygen sensors helps determine converter efficiency.
Answer:
Explanation:
B
How do the particles in plasmas compare with
the particles in solids?
O Plasmas and solids are both made up of cation-anion pairs.
• Solids and plasmas are both made up of electrons and cations.
Solids are made up of cation-anion pairs, but plasmas are not.
O Plasmas are made up of cation-anion pairs, but solids are not.
Answer:
Solids are made up of cation-anion pairs, but plasmas are not
Explanation:
Solid is made from cautions and anions while the plasma is not and hence both are made from the cautions and anion plasma. Solids and plasma is made from electrons and solids are made from caution and anion pairs. Plasma is a good conductor of electricity as they have a lot of mobile charged particles.the ability for carbon to form long chain or rings is
A nuclease enzyme breaks the covalent bond originally connecting the phosphate to the 5' carbon in a nucleic acid. After allowing this enzyme to completely digest the nucleic acid down to monomers, you perform tests to determine where the phosphate is attached to each monomer. Where do you expect to find this phosphate
Answer:
The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
Explanation:
The structure of nucleic acid polymers is built up from monomers of nucleotides.
A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.
When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.
Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
which primitive organic molecule was essential to form lipid bilayer?
a)protenoid
b)phospholipid
c)autocatalytic RNA
d)aminoacids
Answer:
c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.
What is represented by a straight line on a graph?
o the sum of the independent and dependent variables
O only the independent variable
O only the dependent variable
o the relationship between independent and dependent variable
1 2
3
4
5
Answer:
the relationship between independent and dependent variable
Explanation:
A straight line or linear graph is one of the ways to represent a given data. It shows the relationship between two given set of data; one called the independent variable is plotted on the x-axis (horizontal) while the other called the dependent variable is plotted on the y-axis (vertical).
The straighter the line is, the stronger the relationship between the two variables and vice versa. Hence, the straight line in the graph represents the relationship between independent and dependent variable.
The reaction for photosynthesis producing glucose sugar and oxygen gas is:
__CO2(g) + __H2O(l) UV/chlorophyl−→−−−−−−−−−−−−−− __C6H12O6(s) + __O2(g)
What is the volume of oxygen gas at STP produced from 2.20 g of CO2 (44.01 g/mol)?
a. 1.12 L
b. .187 L
c. 4.32 L
d. 6.72 L
e. 1.60 L
Answer:
a. 1.12 L
Explanation:
Step 1: Write the balanced equation for the photosynthesis
6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)
Step 2: Calculate the moles corresponding to 2.20 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
2.20 g × 1 mol/44.01 g = 0.0500 mol
Step 3: Calculate the moles of O₂ produced
The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol
Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP
At STP, 1 mole of O₂ occupies 22.4 L.
0.0500 mol × 22.4 L/1 mol = 1.12 L
How can beta particles be dangerous to living cells?
A. They move fast and penetrate the skin.
B. They are very high in energy and can travel through most
materials.
C. They move slowly but are very large.
D. They are very low in energy but remain in the body for a long time.
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Ladkiyoooooo aaa jaooo♡♥︎♡
Jaldiiiii karo na yaaaar
Answer:
kjajjahahayq :/
Explanation:
a sbywsbgv usnwbhx hg xw nx hb gs
Answer:
Don´t Post Irrelevant Questions!!Explanation:What is the mass of carbon in 69.00 mg of co2
Answer:
18.82 mg
Explanation:
From the given information:
The molar mass of CO2 is calculated as follow
= (12 + (16 ×2))
= 44
The mass of carbon is determined by dividing the mass no of carbon from co2 by the molar mass of CO2, followed by multiplying it by 69.00 mg
= [tex](\dfrac{12}{44}\times 69 )[/tex]
=(0.2727 × 69 )
= 18.82 mg
Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon dioxide. The molar mass of octane is 114.33 g/mole.
2C8H18 + 25O2 → 16 CO2 + 18 H2O
ΔH°rxn = -11018 kJ
Answer:
The right solution is "-602.69 KJ heat".
Explanation:
According to the question,
The 100.0 g of carbon dioxide:
= [tex]\frac{100.0 \ g}{114.33\ g/mole}[/tex]
= [tex]0.8747 \ moles[/tex]
We know that 16 moles of [tex]CO_2[/tex] formation associates with -11018 kJ of heat, then
0.8747 moles [tex]CO_2[/tex] formation associates with,
= [tex]-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat[/tex]
= [tex]-0.0547\times 11018[/tex]
= [tex]-602.69 \ KJ \ heat[/tex]
Congratulations! You are now the head biologist at the local "Cells and Bells" research lab! It has come to other cell biologists' attention recently that some cells are too small to contain all of the organelles inside of them. They decide that it's best to get rid of an organelle, but they're not sure which one. In the first process of this decision, they need to know "which organelle is the most important?"
Your job, as the head cell biologist, is to decide which organelle the cell cannot live without.
Write a research paper (intro, body, and conclusion) on which organelle is the most important and why.
) The C O bond dissociation energy in CO2 is 799 kJ/mol. The maximum wavelength of electromagnetic radiation required to rupture this bond is ________.
Answer:
λ = 150 nm
Explanation:
For C-O bond rupture:
The required energy to rupture C-O bond = bond energy of C-O bond
= 799 kJ/mol
[tex]\mathsf{= 799 \ kJ/mol \times ( \dfrac{1 \ mol }{6.023 \times 10^{23} \ C-O \ bonds })}[/tex]
[tex]\mathsf{= 1.3265 \times 10^{-21} \ kJ/ C-O \ bond}[/tex]
[tex]\mathsf{= 1.33 \times 10^{-18} \ J/C-O \ bond}[/tex]
Recall that the wavelength associated with energy and frequency is expressed as:
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]\lambda = \dfrac{hc}{E}[/tex]
[tex]\lambda = \dfrac{(6.626 \times 10^{-34} \ J.s^{-1}) \times (3.0 \times 10^8 \ ms^{-1})}{ 1.33 \times 10^{-18} \ J/C-O \ bond}}[/tex]
[tex]\mathsf{\lambda = 1.50 \times 10^{-7} \ m}[/tex]
λ = 150 nm
4. A balloon is filled with 3.0 L of helium at 310 K. The balloon is placed in an oven where the
temperature reaches 340 K. What is the new volume of the balloon?
Answer:
3,29L
Explanation:
3.29L = V2
Formula: V1/T1 = V2/T2
--------------------
Given:
V1 = 3.0 L V2 = ?
T1 = 310 K T2 = 340 K
--------------------
Plugin:
(X stands in place of V2 just to make it easier to look at)
[3.0L / 310K = X / 340K]
(3.0L / 310K = 0.01L/K)
0.01L/K = X / 340K
(multiply 340K on both sides, it cancels out on the right)
0.01L/K * 340K = X
(0.01L/K * 340K = 3.29L)
**3.29L = X**
[or]
**3.29L = V2**
Which gas has the highest diffusing rate between nitrogen , oxygen , hydrogen and chlorine
Answer:
hydrogen
Explanation:
The gas with the least molecular weight effuses the fastest (Graham's Law). Hence, H gas has a higher rate of diffusion compared to N, O, and Cl.
So, Cl is the slowest when it comes to the rate of diffusion, because it has the highest molecular weight.
A student has accidentally spilled 100.0 mL of 3.0 mol/L nitric acid onto the lab bench. What mass of sodium bicarbonate would the teacher need to sprinkle on this spill to neutralize and clean it up?
Answer:
25 g
Explanation:
Step 1: Write the balanced equation
HNO₃ + NaHCO₃ ⇒ NaNO₃ + H₂O + CO₂
Step 2: Calculate the reacting moles of HNO₃
100.0 mL of 3.0 mol/L HNO₃ reacted.
0.1000 L × 3.0 mol/L = 0.30 mol
Step 3: Calculate the reacting moles of NaHCO₃
The molar ratio of HNO₃ to NaHCO₃ is 1:1. The reacting moles of NaHCO₃ are 1/1 × 0.30 mol = 0.30 mol.
Step 4: Calculate the mass corresponding to 0.30 moles of NaHCO₃
The molar mass of NaHCO₃ is 84.01 g/mol.
0.30 mol × 84.01 g/mol = 25 g
The half life for the radioactive decay of carbon- to nitrogen- is years. Suppose nuclear chemical analysis shows that there is of nitrogen- for every of carbon- in a certain sample of rock. Calculate the age of the rock. Round your answer to significant digits. g
Answer:
Age of rock = 6.12 × 10³ years
Note: The question is incomplete.A similar but complete question is given below.
The half-life for the radioactive decay of carbon-14 to nitrogen-14 is 5.73 x 10^3 years. Suppose nuclear chemical analysis shows that there is 0.523mmol of nitrogen-14 for every 1.000 mmol of carbon-14 in a certain sample of rock.
Calculate the age of the rock. Round your answer to 2 significant digits.
Explanation:
The half-life of a radioactive material is the time taken for half the atoms in the atomic nucleus of a material to disintegrate.
The half-life for the radioactive decay of carbon-14 to nitrogen-14 is given as 5.73 x 10³ years. This means that given 1 mole of carbon-14 is present initially, after one half-life, 0.5 moles of carbon-14 would remain.
Number of millimoles of carbon-14 remaining = 1 - 0.523 = 0.477 mmol
Number of half-lives that the carbon-14 has undergone is determined as follows:
Amount remaining = (1/2)ⁿ
where nnis number of half-lives
0.5 mmol = one half-life
0.5 = (1/2)¹
O.477 = (1/2)ⁿ = (0.5)ⁿ
㏒₀.₅(0.477) = n
n = ㏒(0.477)/㏒(0.5)
n = 1.067938829
Age of the rock = number of half-lives × half-life
Age of rock = 1.067938829 × 5.73 × 10³ years
Age of rock = 6.12 × 10³ years
NCEPT The mass of NaClcontaining the Avogadro Number of particles is.
Answer:
one mole of water (6.022 x 10 23 molecules) has a mass of 18.02 g. One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.
Explanation:
• The mole (or mol) represents a certain number of objects.
• SI def.: the amount of a substance that contains the same
number of entities as there are atoms in 12 g of carbon-12.
• Exactly 12 g of carbon-12 contains 6.022 x 10 23 atoms.
• One mole of H 2O molecules
contains 6.022 x 10 23 molecules.
• 1 mole contains 6.022 x 10 23 entities (Avogadro’s number)
• One mole of NaCl contains 6.022 x 10 23 NaCl formula units.
• Use the mole quantity to count formulas by weighing them.
• Mass of a mole of particles = mass of 1 particle x 6.022 x 1023
Mass of 1 H atom: 1.008 amu x 1.661 x10-24 g/amu = 1.674 x10-24 g
Mass of 1 mole of H atoms:
1.674 x10-24g/H atom x 6.022 x1023H atoms = 1.008 g
• The mass of an atom in amu is numerically the same
as the mass of one mole of atoms of the element in grams.
• One atom of sulfur has a mass of 32.07 amu;
one mole of S atoms has a mass of 32.07 g
An aqueous solution contains 0.374 M ammonia (NH3). How many mL of 0.276 M nitric acid would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 8.970.
Answer:
111.95mL of HNO3 are needed to prepare the buffer
Explanation:
We can solve this equation using H-H equation for bases:
pOH = pKb + log [HA+] / [A]
Where pOH is the pOH of the solution
pOH = 14 - pH = 14 - 8.970 = 5.03
pKb is the pKb of NH3 = 4.74
[HA+] could be taken as moles of NH4+
[A] as moles of NH3
The NH3 reacts with nitric acid, HNO3, as follows:
NH3 + HNO3 → NH4+ + NO3-
That means the moles of HNO3 added = X = Moles of NH4+ produced
And moles of NH3 are initial moles NH3 - X
Initial moles of NH3 are:
0.125L * (0.374mol/L) = 0.04675 moles NH3
Replacing in H-H equation:
pOH = pKb + log [HA+] / [A]
5.03 = 4.74 + log [X] / [0.04675-X]
0.29 = log [X] / [0.04675-X]
1.95 = [X] / [0.04675-X]
0.0912 - 1.95X = X
0.0912 = 2.95X
X = 0.0309 moles
We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:
0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed
In mL:
111.95mL of HNO3 are needed to prepare the buffer
According to an informal 1992 survey, the drinking water in about one-third of the homes in Chicago had lead levels of about 10 ppb. Dr. Koether lived in Chicago from 1996 to 1998. Assuming she drank 1.4 L of water a day, calculate the total amount of lead in mg (using one decimal place) that she was exposed to over the two years if she lived in a home that had such high levels of lead.
Answer:
10.2 mg
Explanation:
Step 1: Calculate the total amount of water she drank
1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.
If she drank 1.4 L of water per day, the total amount of water she drank is:
730 day × 1.4 L/day = 1022 L
Step 2: Calculate the amount of Pb in 1022 L of water
The concentration of Pb is 10 ppb (10 μg/L).
1022 L × 10 μg/L = 10220 μg
Step 3: Convert 10220 μg to milligrams
We will use the conversion factor 1 mg = 1000 μg.
10220 μg × 1 mg/1000 μg = 10.2 mg
Based on the following observations decide the order of reactivity for hydrogen, magnesium, and copper. Hydrochloric acid reacts with magnesium but did not react with copper. magnesium reacted with copper sulfate. Write your answers in the blanks. For magnesium write magnesium for hydrochloric acid write hydrogen and for copper sulfate write copper.
Answer:
Mg> H> Cu
Explanation:
We can see from the question that hydrochloric acid reacted with magnesium as follows;
Mg(s) + 2HCl(aq) ----> MgCl2(aq) + H2(g)
Copper does not react with HCl which means that copper is less reactive than hydrogen hence it can not displace hydrogen from a dilute acid solution.
The order of reactivity of the elements then is ; Mg> H> Cu
Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How many moles of hydronium ions H3O are neutralized by 1 teaspoon of antacid product?
Answer:
0.0347 moles of hydronium ions
Explanation:
The equation of the neutralization reaction between hydroxide and hydronium ions is given below:
H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)
From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.
The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:
Number of moles = mass / molar mass
Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol
Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol
Mass of magnesium hydroxide = 450 g = 0.45 g
Mass of aluminium hydroxide = 500 mg = 0.5 g
Moles of magnesium hydroxide = (0.45/58) moles
Moles of aluminium hydroxide = (0.5/78) moles
Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:
Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)
Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)
Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles
Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles
Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions
Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.
