Answer:
The correct answer is 0.61 ml
Explanation:
Nitric acid is a strong acid. That means that it dissociates completely in water as follows:
HNO₃ → H⁺ + NO₃⁻
As the dissociation is complete, the concentration of H⁺ ions is equal to the initial concentration of the acid (HNO₃). Thus, the pH can be calculated from the initial concentration of the acid:
pH= -log [H⁺] = -log [acid]
We want a nitric acid solution with a pH of 2.0. so we first calculate the concentration of acid we need:
2.0 = -log [acid]
10⁻²= [acid]
The chemist has a stock solution with C= 9.0 M and he/she wants a solution with C= 1 x 10⁻² M and V= 550 ml. We use the equation that relates the initial concentration and volume (Ci and Vi, respectively) of a solution with the final concentration and volume (Cf and Vf, respectively):
Ci x Vi = Cf x Vf
⇒ Vi= (Cf x Vf)/Ci = (1 x 10⁻² M x 550 ml)/9.0 M = 0.611 ml
Summarizing, the chemist must measure 0.611 ml of concentrated solution (9.0 M), add it to the flask and fill the flask to the mark until 550 ml in order to obtain a nitric acid solution with a pH of 2.0.
The student plans to conduct a spectrophotometric analysis to determine the concentration of Cu2+(aq) in a solution. The solution has a small amount of Co(NO3)2 (aq) present as a contaminant. The student is given a diagram below, which shows the absorbance curves for aqueous solutions of Co2+ (aq) and Cu2+ (aq)
c. The spectrophotometer available to the student has a wavelength range of 400 nm to 700 nm. What wavelength should the student use to minimize the interference from the presence of the Co2+ (aq)?
Answer:
The wavelength the student should use is 700 nm.
Explanation:
Attached below you can find the diagram I found for this question elsewhere.
Because the idea is to minimize the interference of the Co⁺²(aq) species, we should choose a wavelength in which its absorbance is minimum.
At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.
Answer:
Spectrophotometric analysis is a quantitative measure of the interaction of spectrum rays and certain molecules. The wavelength used by student is 700 nanometers.
Explanation:
Spectrophotometric analysis is a method used to measure the amount of absorbance or transmittance done by a substance.
From the knowledge of spectrometric analysis:
Copper is absorbed at 700 nanometers.Cobalt is absorbed between 400-500 nanometers.In the given question, to minimize the interference by cobalt ion, the student should use the wavelength of 700 nanometers or more. The copper is absorbed at wavelength of 700 nm or more, whereas the cobalt ion is absorbed between the range of 400 to 500 nanometres.
Therefore, the student will use wavelength of 700 nanometers because copper is absorbed at this wavelength. The cobalt is absorbed at 400 to 500 nanometers, which will minimize its interference.
For Further Reference:
https://brainly.com/question/22367058?referrer=searchResults
If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate it, what is the acetic acid concentration in the mixture
When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.
Let's consider the neutralization reaction between sodium hydroxide and acetic acid.
NaOH + CH₃COOH ⇒ CH₃COONa + H₂O
32.40 mL of 0.258 M NaOH react. The reacting moles of NaOH are:
[tex]0.03240 L \times \frac{0.258mol}{L} = 8.36 \times 10^{-3} mol[/tex]
The molar ratio of NaOH to CH₃COOH is 1:1. The moles of CH₃COOH required to react with 8.36 × 10⁻³ moles of NaOH are:
[tex]8.36 \times 10^{-3} mol NaOH \times \frac{1molCH_3COOH}{1molNaOH} = 8.36 \times 10^{-3} mol CH_3COOH[/tex]
8.36 × 10⁻³ moles of CH₃COOH are in 1.00 mL of solution. The concentration of CH₃COOH is:
[tex]M = \frac{8.36 \times 10^{-3} mol}{1.00 \times 10^{-3} L} = 8.36 M[/tex]
When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.
Learn more: https://brainly.com/question/2728613
The concentration of acetic acid in the mixture can be found by titrating with NaOH, using a stoichiometric reaction to equate the moles of NaOH to moles of acetic acid, and then calculating the concentration of the acid.
Explanation:This problem is about using acid-base reactions to determine the concentration of acetic acid in a solution. The reaction between the acetic acid and sodium hydroxide (NaOH) is a stoichiometric reaction where one mole of acetic acid reacts with one mole of NaOH. We use that information to calculate the number of moles of acetic acid.
First, we calculate the number of moles of NaOH used in the titration, which is the concentration of NaOH multiplied by the volume used (in liters): moles NaOH = (0.258 mol/L) * (32.40 mL * 1L/1000 mL) = 0.00835 mol NaOH.
Since the reaction is stoichiometric, the number of moles of NaOH is equivalent to the number of moles of acetic acid.
Finally, the concentration of acetic acid is the number of moles divided by the volume of the sample: concentration = moles / volume = 0.00835 mol / 0.001 L = 8.35 M.
Learn more about Acid-Base Titration here:https://brainly.com/question/40172894
#SPJ11
Part 1: Write the complete balanced MOLECULAR equation (including all states of matter) for the precipitation reaction that occurs between aqueous copper(II) chloride and aqueous sodium hydroxide.
Part 2: Write the NET IONIC EQUATION for the REVERSE reaction in Part 1 (include all states of matter).
Part 3: Write the equilibrium expression for the net ionic equation written in Part 2.
Answer:
For 1: The molecular equation is [tex]CuI_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaI(aq)[/tex]
For 2: The net ionic equation is [tex]Cu(OH)_2(s)\rightarrow Cu^{2+}(aq)+2I^{-}(aq)[/tex]
For 3: The equilibrium constant expression is [tex]K_{eq}=[Cu^{2+}][I^-]^2[/tex]
Explanation:
For 1:A molecular equation is defined as the chemical equation in which the ionic compounds are written as molecules rather than component ions.
The molecular equation for the reaction of copper (II) iodide and sodium hydroxide is given as:
[tex]CuI_2(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+2NaI(aq)[/tex]
For 2:Net ionic equation of any reaction does not include any spectator ions.
Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.
The chemical equation for the reaction of copper hydroxide and sodium iodide is given as:
[tex]2NaI(aq)+Cu(OH)_2(s)\rightarrow 2NaOH(aq)+CuI_2(aq)[/tex]
Ionic form of the above equation follows:
[tex]2Na^{+}(aq)+2I^{-}(aq)+Cu(OH)_2(s)\rightarrow 2Na^+(aq)+2OH^-(aq)+Cu^{2+}(aq.)+2I^{-}(aq.)[/tex]
As, sodium and iodide ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.
The net ionic equation for the above reaction follows:
[tex]Cu(OH)_2(s)\rightarrow Cu^{2+}(aq)+2I^{-}(aq)[/tex]
For 3:The expression of equilibrium constant for the net ionic equation above follows:
[tex]K_{eq}=[Cu^{2+}][I^-]^2[/tex]
Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.
Can a metal and a nonmetal participate in a combination reaction
Answer:
Yes
Explanation:
A reaction normally takes place between metals and non metals. The metals acts as the electron donors and the non metals acts as the electron acceptors. This exchange of electrons form bonds such as ionic or covalent.
A good example of a reaction between a metal and non metal is Sodium metal and Chlorine(non metal). They form an ionic bond and the product is Sodium chloride.
Calculate the percent dissociation of trimethylacetic acid in a aqueous solution of the stuff. You may find some useful data in the ALEKS Data resource. Round your answer to significant digits.
Final answer:
To calculate the percent dissociation of trimethylacetic acid in an aqueous solution, use the equilibrium constant expression and the initial concentration of the acid.
Explanation:
To calculate the percent dissociation of trimethylacetic acid in an aqueous solution, we need to use the equilibrium constant expression and the initial concentration of the acid. The equilibrium expression for the dissociation of trimethylacetic acid is written as:
Ka = [CH₃COOH]/[CH₃C₃H₇OH]
where [CH₃COOH] represents the concentration of trimethylacetic acid and [CH₃C₃H₇OH] represents the concentration of the dissociated form of the acid.
By rearranging the equation, we can solve for the percent dissociation:
Percent Dissociation = ([CH₃C₃H₇OH]/[CH₃COOH]) x 100
Using the given information or data from the ALEKS Data resource, we can substitute the values into the equation and calculate the percent dissociation.
There is a lot of talk about pH and shampoos. Some shampoos claim they balance the pH in your hair, others say they are alkaline and others say they are neutral or match the "natural" pH of your hair. What do you think that this means? Do you think the pH of your shampoo affects the condition of your hair? Why or why not? *
Answer:Yes it affects the conditions of the hair.
Explanation:most shampoos have pH values higher than the pH of the hair shaft of 3.6 and also higher than the pH of the hair scalp of 5.5. Therefore the usage of hair shampoo with pH values above 5.5 may likely increase the rate of friction which may cause hair breakage and facilitate hair tangling.
Shampoos with various pH ratings have different effects on hair. The pH level of your shampoo influences the health of your hair. Shampoos that match the hair's natural pH of 5 to 7 are typically best for maintaining hair health.
Explanation:The pH measurement, which ranges from 0 to 14, indicates whether a substance is acidic, neutral, or alkaline. Most shampoos usually have a pH level between 5 and 7, similar to hair's natural pH. When a shampoo claims to 'balance' the pH of your hair, it means it tries to keep this pH level, which is slightly acidic for a healthy scalp and hair. Meanwhile, neutral shampoos have a pH of 7, and alkaline shampoos have a pH higher than 7, which could potentially cause hair to become dry and brittle over time, especially with overuse. So, the pH of your shampoo does indeed affect your hair condition. Using a shampoo that aligns with the natural pH of your hair is generally considered optimal for maintaining hair health.
Learn more about shampoo pH here:https://brainly.com/question/32512053
#SPJ6
It is often possible to change a hydrate into an anhydrous compound by heating it to drive off the water (dehydration). Write an equation that shows the dehydration of manganese(II) sulfate monohydrate .
Answer:
Yes is possible
Explanation:
MnSO4*7H2O(s) → MnSO4(s) + 7H2O(g)
This should be an endothermic reaction. The water molecules are interacting with the Mn2+ and SO42- ions within the crystal lattice of the compound. You must overcome those interactions, and so that would require the input of energy.
One mole of an atom of a substance is equal to
Answer:
One mole of a substance is equal to 6.022 × 10²³ units of that substance (atoms, molecules, or ions)
Explanation:
This number is Avogadro's number. The concept of the mole can be used to convert between mass and number of particles. its used to compare very large numbers.
Water standing in the open at 33.0°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (557 cal/g) is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per gram. (a) Find ε. (b) What is the ratio of ε to the average kinetic energy of H2O molecules, assuming the latter is related to temperature in the same way as it is for gases?
Answer:
a) ε = 6.961 × 10⁻²⁰Joules
b)The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642
Explanation:
a) The formula to be used is given below as :
Heat of Vapourisation(Lv) = εn
Where: ε = is the average energy of the escaping molecules and
n = is the number of molecules per gram
The first step is to convert 557 cal/g to joules/kilogram(j/kg)
1 cal/g = 4186.8j/kg
557cal/g = ?
We cross multiply
557cal/g × 4186.8j/kg
= 2332047.6j/kg
Therefore, 557 cal/g = 2332047.6j/kg
ε = Lv/n
ε = LvM/n
Where Lv = 2332047.6j/kg
M = 0.018kg/mol
n = 6.03 × 10²³mol
ε =( 2332047.6j/kg × 0.018kg/mol) ÷ 6.03 × 10²³mol
= 6.961336119 × 10⁻²⁰Joules
Approximately, ε = 6.961 × 10⁻²⁰Joules
b) Kinetic energy = (3/2)KT
The ratio of ε to Kinetic energy = ε/(3/2)kT = 2ε /3kT
Where ε = 6.961 × 10⁻²⁰Joules
k = 1.38× 10⁻²³ Joules/kelvin
T = 33°C , which will be converted to kelvin as
33°C + 273K
= 306K
The ratio of ε to Kinetic energy will be calculated as
2ε /3kT
= (2×6.961 × 10⁻²⁰ Joules) ÷ (3 × 1.38× 10⁻²³Joules/kelvin × 306K)
= 10.642
Hence , The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642
Select each of the reactions and observe the reactants taking part in the reactions. Suppose you are carrying out each of these reactions starting with five moles of each reactant. Observe that some reactants are in excess whereas some reactants limit the amount of products formed.
Classify each of the reactants as a limiting reactant or an excess reactant for a reaction starting with five moles of each reactant.
A. H2 in formation of water
B. O2 in formation of water
C. CH4 in combustion of methane
D. N2 in formation of ammonia
E. H2 in formation of ammonia
F. O2 in combustion of methane
Answer:
A. In the formation of water, H2 is the limiting reactant and O2 is the excess reactant.
B. In the Combustion of methane (CH4), O2 is the limiting reactant and CH4 is the excess reactant.
C. In the formation of ammonia (NH3), H2 is the limiting reactant and N2 is the excess reactant.
Explanation:
The question suggests that each reactant has 5 moles.
A. Formation of water. Water is produced when H2 and O2 combine together according to the balanced equation below:
2H2 + O2 —> 2H2O
From the balanced equation above,
2 moles of H2 reacted with 1 mole of O2.
Therefore, 5 moles of H2 will react with = 5/2 = 2.5 moles of O2.
From the above calculations, we can see that there are left over for O2 as only 2.5 moles reacted out of the 5 moles that was given.
Therefore, H2 is the limiting reactant and O2 is the excess reactant.
B. Combustion of methane. Combustion is simply a reaction in the presence of oxygen. The balance equation for the Combustion of methane (CH4) is given below:
CH4 + 2O2 —> CO2 + 2H2O
From the balanced equation above,
1 mole of CH4 reacted with 2 moles of O2.
Therefore, 5 moles of CH4 will react with = 5 x 2 = 10 moles of O2.
From the calculations made above, we can see that it requires higher amount of O2 to react with 5 moles CH4. Therefore, O2 is the limiting reactant and CH4 is the excess reactant.
C. Formation of ammonia.
Ammonia is obtained when N2 and H2 combine according to the balanced equation below:
N2 + 3H2 —> 2NH3
From the balanced equation above,
1 mole of N2 reacted with 3 moles of H2.
Therefore, 5 moles of N2 will react with = 5 x 3 = 15 moles of H2.
From the calculations made above, a higher amount of H2 is required to react with 5 moles of N2. Therefore, H2 is the limiting reactant and N2 is the excess reactant.
Final answer:
To determine limiting or excess reactants, compare the stoichiometry of the balanced equations to the amounts provided. For reactions involving the formation of water or ammonia, and the combustion of methane, the limiting reactant is based on the stoichiometric ratio required for each reactant. Hence, O2 typically ends up as the limiting reactant for the formation of water and combustion of methane, while N2 is the limiting reactant in the formation of ammonia.
Explanation:
When determining whether a reactant is a limiting reactant or an excess reactant, you must look at the balanced chemical equations for the reactions. Consider the stoichiometric coefficients, which tell you the proportions in which reactants combine to form products. If starting with five moles of each reactant:
A. H2 in formation of water: The balanced equation is 2H2 + O2To identify the limiting reactant, compare the molar ratios of the reactants to those in the balanced equation. The reactant that is in a smaller proportion relative to its stoichiometric coefficient in the equation is the limiting reactant. The one in the larger proportion is the excess reactant.
Which of the following has the most kinetic energy?
A car traveling at 80 kph. incorrect answer
B tractor-trailer traveling at 80 kph. incorrect answer
C cheetah running at 80 kph. incorrect answer
D motorcycle traveling at 80 kph.
Answer:
Motorcycle
Explanation:
If you have 3kg of lead and put 3840 joules of heat energy into that mass of lead. Taking
careful measurements, you observe the temperature of the lead rise from 40C to 50C.
What is the specific heat capacity of lead?
Answer : The specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]
Explanation :
Formula used:
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = 3840 J
m = mass of lead = 3 kg = 3000 g
c = specific heat capacity of lead = ?
[tex]T_1[/tex] = initial temperature = [tex]40^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]50^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]3840J=3000g\times c\times (50-40)^oC[/tex]
[tex]c=0.128J/g^oC[/tex]
Therefore, the specific heat capacity of lead is, [tex]0.128J/g^oC[/tex]
3. The following reaction reaches equilibrium at 650 ˚C and atmospheric pressure: N2 (g) + C2H2 (g) 2HCN (g) If the system initially is an equimolar mixture of nitrogen and acetylene, what is the composition of the system at equilibrium? What would be the effect of doubling the pressure? Assume ideal gases.
At equilibrium, the composition of the system is determined by the stoichiometry of the reaction. Doubling the pressure will not have a significant effect on the composition of the system at equilibrium.
Explanation:At equilibrium, the composition of the system can be determined using the stoichiometry of the reaction. In this reaction, one N2 molecule reacts with one C2H2 molecule to produce two HCN molecules. Therefore, at equilibrium, the concentration of N2 and C2H2 will both decrease by the same amount, while the concentration of HCN will increase by twice that amount.
If the pressure is doubled, the system will try to relieve the stress by shifting the equilibrium to the side with fewer moles of gas. In this case, the total number of moles of gas will remain the same, so doubling the pressure will not have a significant effect on the composition of the system at equilibrium.
Learn more about Equilibrium Composition here:https://brainly.com/question/30310248
#SPJ12
Select the correct value for the indicated bond angle in each of the compounds.
90°, 180°, 109.5°, 120°, <120°, <109.5°
a) O−O−O angle of O3
b) F−B−F angle of BF3
c) F−O−F angle of OF2
d) Cl−Be−Cl angle of BeCl2
e) F−P−F angle of PF3
f) Cl−Si−Cl angle of SiCl4
The bond angles for ozone (O₃), boron trifluoride (BF₃), oxygen difluoride (OF₂), beryllium dichloride (BeCl₂), phosphorus trifluoride (PF₃), and silicon tetrachloride (SiCl₄) are <120°, 120°, <109.5°, 180°, <109.5°, and 109.5° respectively, based on their molecular geometries.
The correct value for the indicated bond angle in each of the compounds is based on the molecular geometry and electron-pair repulsion theory. Here are the answers:
Ozone (O₃): The O-O-O angle in O₃ is expected to be <120° due to its bent molecular structure.
Boron trifluoride (BF₃): The F-B-F angle in BF3 is 120° because of its trigonal planar molecular geometry.
Oxygen difluoride (OF₂): The F-O-F angle in OF2 is expected to be <109.5° because fluoro substituents cause greater repulsion than hydrogens, narrowing the bond angle from the tetrahedral
Beryllium dichloride (BeCl₂): The Cl-Be-Cl angle in BeCl₂ is 180° due to its linear molecular structure.
Phosphorus trifluoride (PF₃): The F-P-F angle in PF₃ is expected to be <109.5° due to its trigonal pyramidal molecular structure.
Silicon tetrachloride (SiCl₄): The Cl-Si-Cl angle in SiCl₄ is 109.5° since it has a tetrahedral molecular geometry.
what statement correctly
describe the nucleus of an atom
A. the nucleus of the atom contains the electrons
B. the nucleus of the atom is mostly empty space
C. the nucleus of the atom surrounds the center of the atom
D. the nucleus of the atom contains most of the mass of an atom
The nucleus of an atom, comprised of protons and neutrons, contains most of the atom's mass. Despite its relatively minuscule size, it is much heavier than the electron cloud surrounding it, which occupies most of the atom's volume.
Explanation:The statement that correctly describes the nucleus of an atom is: D. the nucleus of the atom contains most of the mass of an atom.
Modern atomic theory has shown us that the nucleus of an atom is made up of positively charged protons and uncharged neutrons, and overall, carries a positive charge. The nucleus is incredibly tiny, with its diameter being about 100,000 times smaller than the diameter of the atom. However, it contains almost all of the atom's mass as protons and neutrons are significantly heavier than electrons, thus making statement D accurate.
Electrons, on the other hand, occupy the majority of an atom's volume, constituting an 'electron cloud' around the nucleus. Because they are very lightweight, their specific locations can't be definitively pinpointed and are often represented as probabilities within this cloud. Hence, options A, B, and C are incorrect.
Learn more about Nucleus of an Atom here:https://brainly.com/question/10658589
#SPJ12
2) A professor wanted to set up a similar experiment as the one you performed in lab. He wanted to use Al(OH)3 in place of Ca(OH)2. Calculate how many mL of saturated Al(OH)3 solution it would take to titrate against 12.00 mL of 0.0542 M HCl solution. The Ksp of Al(OH)3 is 3.0x10-34. Show your work to receive credit. Finally, do you think this would be reasonable experiment for a general chemistry lab
Explanation:
The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.
This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated
The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.
The isomerization of methylisonitrile to acetonitrile (CH3NC(g) ???? CH3CN) is first order in CH3NC. The rate constant for the reaction is 9.45 x 10-5 s-1 at 478 K. What is the half-life of the reaction when the initial concentration of CH3NC is 0.0300 M?
Answer:
The half life is [tex]H_{1/2}= 7333.3sec[/tex]
Explanation:
The half life of a first order reaction is mathematically represented as
[tex]H_{1/2} = \frac{0.693}{Rate Constant }[/tex]
Substituting [tex]9.45 * 10^{-5}s^{-1}[/tex] for the rate constant
[tex]H_{1/2} = \frac{0.693}{9.45*10^{-5}}[/tex]
[tex]H_{1/2}= 7333.3sec[/tex]
Final answer:
The half-life of the isomerization of methylisonitrile to acetonitrile, which is a first-order reaction with a rate constant of 9.45 x 10⁻⁵ s⁻¹, is approximately 7330 seconds.
Explanation:
The student's question asks for the calculation of the half-life of the reaction where methylisonitrile isomerizes to acetonitrile (CH₃NC to CH₃CN). This reaction is first order with respect to methylisonitrile, and the rate constant k is given as 9.45 x 10⁻⁵ s⁻¹ at 478 K. To find the half-life ([tex]t^{1/2[/tex]), we can use the first-order half-life equation, [tex]t^{1/2[/tex] = (ln 2) / k. Substituting the given rate constant into this equation
[tex]t^{1/2[/tex] = (ln 2) / (9.45 x 10⁻⁵ s⁻¹)
[tex]t^{1/2[/tex] = (0.693) / (9.45 x 10⁻⁵ s⁻¹)
[tex]t^{1/2[/tex] ≈ 7330 seconds
The half-life of the reaction under the given conditions is approximately 7330 seconds.
A student may say hypothesis that the water on the outside of the paint evaporate faster than the inside . Describe an investigation of the student could use to test this hypothesis
Answer:
First of all, exterior paint is more resistant to extreme temperatures, humidity, pressure differences, the presence of droughts and this is due to its chemical composition, which is more resistant and because it penetrates the surface to be painted better (than in This case is the external wall of a home), by better penetrating the surface to which it is applied, there is less risk of detachment, of accumulating bubbles or air at the wall-paint interface, it adheres much better ... as explained this? Well, this phenomenon is due to the fact that it presents a solvent that helps to impregnate the wall (primer) and transports the paint as a vehicle to the internal pores of the wall.
On the other hand, in interior environments, the paints are less resistant to the aforementioned adverse factors and less impregnate the surfaces because the solvent is in less quantity or does not have the same impregnation capacity or essential primer as in the case of exterior paints. .
Explanation:
There are some specialists in the field who nowadays advise even interior paints with insulating or waterproof lacquers since the interior ones bear so little factors such as humidity that they no longer meet the basic requirements, that is why some interior paints they show fungus or wet stains against walls with wet content.
Answer:
Water on the outside of the paint evaporate faster than the inside. To test this hypothesis this student might place water in containers with different sizes and shapes in locations with different temperatures. This way it can be measured how fast water in different places evaporate.
Explanation:
A change in temperature can affect the physical properties of water. Water circulates through a continuous cycle. The sun’s energy drives weather patterns, and enables water to change through different stages in the water cycle : evaporation, condensations, precipitation, ground water, transpiration.
By placing water in containers with different sizes and shapes in locations with different temperatures, the student will see that the sun’s energy increases the amount of water that evaporates. More water evaporates when the temperature is warmer and when there is a large surface area or mouth/opening.
The expression below was formed by combining different gas laws.
Van
Which law was used to determine the relationship between the volume and the number of moles in this equation?
Boyle's law
Charles's law
Avogadro's law
Gay-Lussac's law
The law used to determine the relationship between the volume and the number of moles in this equation is called the Avogadro's law. by this law, the volume of gas is directly proportional to the number of moles of the gas.
What is Avogadro's law ?Avogadro's law, which was proposed by the Italian scientist Amedeo Avogadro, explains the behavior of gases and states that gases of different chemical nature and physical conditions at the same temperature and pressure have an equal number of molecules in equal volumes.
In simpler terms, the number of molecules or moles in a gas sample is directly proportional to its volume at a constant temperature and pressure.
This law is significant in various areas of chemistry, including the calculation of chemical reactions involving gases, determination of molar volumes, and the study of ideal gases.
Find more on Avogadro's law :
https://brainly.com/question/6534668
#SPJ7
Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that of hydrofluorocarbons. One source of NO· in the stratosphere is supersonic aircraft whose jet engines convert small amounts of N2 and O2 to NO·. Write the propagation steps for the reaction of O3 with NO·.
Answer:
O3 + NO. = O2. + NO2
2O. +NO2 = NO. + 2O2
Explanation:
NO. = Radical
O. & O2. = Radical
Final answer:
Nitric oxide (NO) catalytically destroys ozone (O3) in the stratosphere via a series of propagation steps, contributing to ozone layer depletion and increasing exposure to harmful UV rays.
Explanation:
Nitric oxide (NO) plays a significant role in the destruction of ozone (O3) through a series of reactions in the stratosphere. The presence of NO contributes to the depletion of the ozone layer, which can increase the risk of harmful UV exposure to humans and the environment. The relevant ozone depletion mechanism proceeds with nitric oxide acting as a catalyst through the following propagation steps:
NO + O3 → NO2 + O2NO2 + O3 → NO3 + O2NO3 + NO2 → N2O5These reactions show how NO continues to regenerate and is not consumed, which is why it acts as a catalyst, increasing the rate of ozone decomposition. The nitric oxide is primarily produced by high-flying aircraft and motor vehicles, raising concerns about its environmental impact.
Consider the reversible reaction.
PCl5↽−−⇀PCl3+Cl2PCl5↽−−⇀PCl3+Cl2
Are the concentrations of phosphorus pentachloride, PCl5,PCl5, and phosphosphorus trichloride, PCl3,PCl3, constant or changing at equilibrium?
a.The concentration of PCl5PCl5 is constant, and the concentration of PCl3PCl3 is changing at equilibrium.
b.The concentration of PCl5PCl5 is changing, and the concentration of PCl3PCl3 is constant at equilibrium.
c.The concentrations of both PCl5PCl5 and PCl3PCl3 are constant at equilibrium.
d.The concentrations of both PCl5PCl5 and PCl3PCl3 are changing at equilibrium.
which one is correct please explain
In a reversible reaction at equilibrium, including the reaction PCl5↽−−⇀PCl3+Cl2, the concentrations of all components, including PCl5 and PCl3, remain constant because the forward and reverse reactions occur at the same rate.
Explanation:In a reversible reaction at equilibrium the concentrations of all reactants and products remain constant. This is because the rate of the forward reaction equals the rate of the reverse reaction. So, in the context of the provided equation PCl5↽−−⇀PCl3+Cl2, when the system reaches equilibrium, no changes are noticeable, meaning the concentrations of phosphorus pentachloride, PCl5, and phosphorus trichloride, PCl3, as well as chlorine (Cl2), are constant. Therefore, the correct answer is (c) The concentrations of both PCl5 and PCl3 are constant at equilibrium.
Learn more about Chemical Equilibrium here:https://brainly.com/question/3920294
#SPJ12
At equilibrium, the concentrations of both phosphorus pentachloride (PCl5) and phosphorus trichloride (PCl3) are constant due to the condition of dynamic equilibrium. The reaction continues to occur, but the amounts of reactants and products remain unchanged.
Explanation:In the context of the reversible reaction PCl5 ↔ PCl3 + Cl2, at equilibrium, both the concentration of phosphorus pentachloride, PCl5, and the concentration of phosphorus trichloride, PCl3, are constant.
This is described as a condition known as dynamic equilibrium, where the forward and reverse reactions are happening at the same rate, meaning the concentrations of PCl5, PCl3, and Cl2 are not changing, even though the reacting process is still happening. Therefore, the correct answer is (c) The concentrations of both PCl5 and PCl3 are constant at equilibrium.
Learn more about Dynamic Equilibrium here:https://brainly.com/question/26034510
#SPJ2
the internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure.
Answer:
Check Explanation
Explanation:
The Mathematical expression of the first law is given as
ΔU = Q - W
Note that depending on convention, the mathematical expression for the first law can be written as
ΔU = Q + W
a) Yes, the gas does work on its surroundings. Since it expands; indicating that there is a change in volume.
The work done is given as -PΔV if we are using the convention of (ΔU = Q + W). And this work is negative work done, since the system expands and does work on the surroundings.
But if we are using the convention of (ΔU = Q - W), the work done is given as PΔV and the workdone by the system on the environment is positive work during expansion.
b) Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.
So, from the mathematical expression,
ΔU = Q + W or ΔU = Q - W
If ΔU = 0,
Q = - W or Q = W
But either ways,
Q = PΔV
(because, W = -PΔV for ΔU=Q+W and W = PΔV for ΔU=Q-W)
So, either ways, the heat transfer is the same and it is positive. This indicates heat is transferred from the surroundings to the system.
(c) What is ΔU for the gas for this process?
Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.
Hope this Helps!!!
Based on the given question, we can state that Yes, the gas does work on its surroundings because it expands, this shows that there is a change in volume.
Based on the second question, we can state that because this takes place at constant temperature and against a constant atmospheric pressure, then the change in internal energy for this system is zero.
Based on the third question, we can see that the ΔU for the gas for this process is zero because of the constant temperature and atmospheric pressure.
The Mathematical expression of the first lawThis is given as
ΔU = Q - W
Hence, from the mathematical expression,
ΔU = Q + W or ΔU = Q - WIf ΔU = 0,Q = - W or Q = WRegardless,
Q = PΔV
This means that the heat transfer remains the same and is positive
Read more about heat transfer here:
https://brainly.com/question/12072129
Tooth enamel is composed of hydroxyapatite, Ca5(PO4)3OH (Ksp = 6.8 10-37). Many water treatment plants add F- ion to drinking water, which reacts with Ca5(PO4)3OH to form the more tooth decay-resistant fluorapatite, Ca5(PO4)3F (Ksp = 1.0 10-60). Fluoridated water has resulted in a dramatic decrease in the number of cavities among children. Calculate the solubility of Ca5(PO4)3F in water.
Answer:
6.1 × 10⁻⁸ M
Explanation:
Let's consider the solution of fluorapatite.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
We can relate the solubility (S) with the solubility product constant (Ksp) using an ICE chart.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
I 0 0 0
C +5 S +3 S +S
E 5 S 3 S S
The Ksp is:
Ksp = [Ca²⁺]⁵ × [PO₄³⁻]³ × [F⁻] = (5 S)⁵ × (3 S)³ × S = 84,375 S⁹
[tex]S = \sqrt[9]{\frac{Ksp}{84,375} } = \sqrt[9]{\frac{1.0 \times 10^{-60} }{84,375} } =6.1 \times 10^{-8} M[/tex]
The solubility of fluorapatite compound Ca₅(PO₄)₃F in water is 6.1 × 10⁻⁸M.
What is Ksp?Ksp is the solubility product constant which tells about the relative solubility of the compound which is in equilibrium with their constitute ions.
Chemical reaction for the solubility of fluorapatite compound is:
Ca₅(PO₄)₃F(s) ⇄ 5Ca²⁺(aq) + 3PO₄³⁻(aq) + F⁻(aq)
Let at equilibrium formed concentration of Ca²⁺, PO₄³⁻ and F⁻ are 5x, 3x and x respectively.
Given value of Ksp for Ca₅(PO₄)₃F = 1.0 × 10⁻⁶⁰
Ksp equation for the given reaction will be represented as:
Ksp = [Ca²⁺]⁵[PO₄³⁻]³[F⁻]
On putting above values in the equation we get,
1.0 × 10⁻⁶⁰ = (5x)⁵. (3x)³. (x)
x = 6.1 × 10⁻⁸M
Hence, solubility of Ca₅(PO₄)₃F is 6.1 × 10⁻⁸M.
To know more about Ksp, visit the below link:
https://brainly.com/question/7185695
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of wate
Answer: The final temperature of the mixture is 49°C
Explanation:
To calculate the mass of water, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
For cold water:Density of cold water = 1 g/mL
Volume of cold water = 230.0 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g[/tex]
For hot water:Density of hot water = 1 g/mL
Volume of hot water = 120.0 mL
Putting values in above equation, we get:
[tex]1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g[/tex]
When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of hot water = 120 g
[tex]m_2[/tex] = mass of cold water = 230 g
[tex]T_{final}[/tex] = final temperature = ?°C
[tex]T_1[/tex] = initial temperature of hot water = 95°C
[tex]T_2[/tex] = initial temperature of cold water = 25°C
c = specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
[tex]120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)][/tex]
[tex]T_{final}=49^oC[/tex]
Hence, the final temperature of the mixture is 49°C
Steam reforming of methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills 1.5L a flask with of methane gas and 2.6 atm of water vapor at 47C, and when the mixture has come to equilibrium measures the partial pressure of carbon monoxide gas to be 1.4 atm.
Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
Answer:
Answer: Kp = 4.5
Explanation:
The balanced equation for the production of synthetic gas is
CH4 + H20 ⇒ CO + 3H2
Let x be the change in the concentration of each species at equilibrium
CH4 + H20 ⇔ CO + 3H2
Initial 0.60 2.6 0 0
Change -x -x +x +3x
Equilibrium 0.60-x 2.6-x x 3x
Given that the equilibrium partial pressure of H2= 1.4 atm
then, 3x= 1.40
x= 1.4/3 = 0.466667
The equilibrium concentrations are
{CH4} = 0.60- x = 0.60 - 0.466667 = 0.133333atm
{H2O} = 2.60- x = 2.60 - 0.466667 = 2.133333atm
{CO} = x = 0.466667atm
{H2} = 1.4atm (given)
Kp = {CO}{H2}³
{CH4}{H2O}
Kp = (0.466667)(1.4)³
(0.133333)(2.133333)
= 4.501875
Kp = 4.5
Calculate the pH for each case in the titration of 50.0 mL of 0.110 M HClO ( aq ) with 0.110 M KOH ( aq ) . Use the ionization constant for HClO . What is the pH before addition of any KOH ? pH = What is the pH after addition of 25.0 mL KOH ? pH = What is the pH after addition of 35.0 mL KOH ? pH = What is the pH after addition of 50.0 mL KOH ? pH = What is the pH after addition of 60.0 mL KOH ? pH =
pH before KOH: 4.23. After 25.0 mL KOH: 4.23. After 35.0 mL: 9.98. After 50.0 mL: 11.26. After 60.0 mL: 11.85.
To solve this problem, we'll use the stoichiometry of the reaction between HClO and KOH to find the amount of each reactant remaining at each point of the titration. Then, we'll use the relevant equilibrium equations to calculate the pH.
Step 1: Calculate the initial pH before adding any KOH.
[tex]\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \][/tex]
[tex]\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{K_a [\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{(3.5 \times 10^{-8})(0.110)} \][/tex]
[tex]\[ [\text{H}^+] \approx 5.92 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ \text{pH} = -\log{[\text{H}^+]} = -\log{(5.92 \times 10^{-5})} \][/tex]
[tex]\[ \text{pH} \approx 4.23 \][/tex]
Step 2: Calculate the pH after adding 25.0 mL of KOH.
Moles of HClO initially = [tex]\( 0.110 \, \text{M} \times 0.0500 \, \text{L} = 0.00550 \, \text{mol} \)[/tex]
Moles of KOH added = [tex]\( 0.110 \, \text{M} \times 0.0250 \, \text{L} = 0.00275 \, \text{mol} \)[/tex]
Remaining moles of HClO = [tex]\( 0.00550 \, \text{mol} - 0.00275 \, \text{mol} = 0.00275 \, \text{mol} \)[/tex]
Calculate the new concentration of HClO:
[tex]\[ \text{New volume of HClO} = 50.0 \, \text{mL} - 25.0 \, \text{mL} = 25.0 \, \text{mL} = 0.0250 \, \text{L} \][/tex]
[tex]\[ \text{New} \, [\text{HClO}] = \frac{0.00275 \, \text{mol}}{0.0250 \, \text{L}} \][/tex]
[tex]\[ \text{New} \, [\text{HClO}] = 0.110 \, \text{M} \][/tex]
Now, we use the equilibrium equation for the dissociation of HClO to find the concentration of [tex]\( \text{H}^+ \):[/tex]
[tex]\[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \][/tex]
[tex]\[ K_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{K_a [\text{HClO}]} \][/tex]
[tex]\[ [\text{H}^+] = \sqrt{(3.5 \times 10^{-8})(0.110)} \][/tex]
[tex]\[ [\text{H}^+] \approx 5.92 \times 10^{-5} \, \text{M} \][/tex]
[tex]\[ \text{pH} = -\log{[\text{H}^+]} = -\log{(5.92 \times 10^{-5})} \][/tex]
[tex]\[ \text{pH} \approx 4.23 \][/tex]
The pH remains the same after adding 25.0 mL of KOH.
Let's now calculate the pH after adding 35.0 mL, 50.0 mL, and 60.0 mL of KOH.
9) How long would it take 100 molecules of valinomycin to transport enough K+ to change the concentration inside an erythrocyteof volume 100um3by10 mM? (Assume that valinomycin does not also transport any K+ out of the cell, which it really does, and that valinomycin molecules outside the cell are always saturated with K+.)
Answer: Time required for 100 molecules of valinomycin to transport enough [tex]K^{+}[/tex] to change the concentration inside an erythrocyte is 10 minutes.
Explanation:
The given data is as follows.
Ions to be transported = [tex](10 mM) \times (100 \mu^{3})(N)[/tex]
= [tex]0.010 mol \times 10^{-13} L \times 6.02 \times 10^{23} ions/mol[/tex]
= [tex]6.02 \times 10^{6}[/tex] ions
Here, there are 100 ionophores are present and each one of ionophores are required to transport a value of ions.
Therefore, the time required for 100 molecules of valinomycin to transport enough [tex]K^{+}[/tex] to change the concentration inside an erythrocyte is as follows.
[tex]6.02 \times 10^{6} ions \times \frac{1 sec}{10^{4} ions}[/tex]
= 602 sec
or, = 10 minutes (as 1 min = 60 sec)
Thus, we can conclude that time required for 100 molecules of valinomycin to transport enough [tex]K^{+}[/tex] to change the concentration inside an erythrocyte is 10 minutes.
A bomb calorimeter was used to measure the heat of combustion of naphthalene (C10H8 ). The temperature of the water rose from 25.00°C to 30.70°C. If 1.44 g of naphthalene was used, what is the heat of combustion of one mole of naphthalene? The heat capacity of the calorimeter and its surroundings is 10.17 kJ/°C.
Final answer:
The heat of combustion of one mole of naphthalene is 591 kJ/mol.
Explanation:
To calculate the heat of combustion of one mole of naphthalene, we first need to calculate the heat released by the combustion of the given mass of naphthalene. We can use the formula q = mcΔT, where q is the heat released, m is the mass of the naphthalene, c is the heat capacity of the calorimeter and its surroundings, and ΔT is the change in temperature. Substituting the given values into the equation, we have:
q = 1.44 g * 10.17 kJ/°C * (30.70°C - 25.00°C)
q = 1.44 g * 10.17 kJ/°C * 5.7°C
q = 82.9728 kJ
Now, to calculate the heat of combustion per mole of naphthalene, we will use the molar mass of naphthalene, which is 128.18 g/mol:
Heat of combustion per mole = 82.9728 kJ / 1.44 g * 128.18 g/mol
Heat of combustion per mole = 591 kJ/mol
Final answer:
To determine the heat of combustion per mole of naphthalene, the heat released was first calculated using the calorimeter's heat capacity and the observed temperature change. This amount was then divided by the mass of naphthalene burned to find the heat of combustion per gram, which was finally multiplied by the molar mass of naphthalene to get the value per mole.
Explanation:
To find the heat of combustion of one mole of naphthalene using a bomb calorimeter, we first need to use the given heat capacity of the calorimeter, which is 10.17 kJ/°C, and the temperature change that the combustion caused in the water, which was from 25.00°C to 30.70°C. The mass of naphthalene combusted was 1.44 g.
First, calculate the total heat q released during the combustion:
q = C₁₄₂₃ bomb × ΔT = (10.17 kJ/°C) × (30.70°C - 25.00°C) = 10.17 kJ/°C × 5.70°C = 57.969 kJ
Next, calculate the heat of combustion per gram of naphthalene:
q per gram = 57.969 kJ / 1.44 g = 40.2604 kJ/g
Now, to find the heat of combustion per mole of naphthalene, we multiply this value by the molar mass of naphthalene. The molar mass of naphthalene (C₁₀H₈) is approximately 128.17 g/mol.
Heat of combustion per mole = q per gram × Molar Mass = 40.2604 kJ/g × 128.17 g/mol ≈ 5160.76 kJ/mol
Therefore, the heat of combustion of one mole of naphthalene is around 5160.76 kJ/mol.
Propane (C3H8) burns in oxygen to produce carbon dioxide gas and water vapor. (a) Write a balanced equation for this reaction. Include all reaction states. → (b) Calculate the number of liters of carbon dioxide measured at STP that could be produced from 7.65 g of propane. L
Answer:
11.6 L will be the number of liters of carbon dioxide measured at STP.
Explanation:
The balanced equation for this combustion reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
where 1 mol of propane reacts to 5 moles of oxygen in order to produce 3 moles of carbon dioxide and 4 moles of water.
We assume the oxygen in excess, so the limiting reagent is the propane. Now, we determine the moles: 7.65 g . 1 mol/ 44 g = 0.174 moles
Ratio is 1:3. 1 mol of propane can produce 3 moles of CO₂
Therefore, 0.174 moles will produce (0.174 . 3) / 1 = 0.521 moles of CO₂
As 1 mol of gas is contained in 22.4L at STP conditions, we propose
22.4L / 1 mol = V₂ / 0.521 mol
22.4 L / 1 mol . 0.521 mol = V₂ → 11.6 L
Answer:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)
11.6 L of CO2 will be produced
Explanation:
Step 1: Data given
Mass of propane = 7.65 grams
Molar mass propane = 44.1 g/mol
Burning = combustion reation = adding O2. The products will be CO2 and H2O
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g)
Step 3: Calculate moles propane
Moles propane = 7.65 grams / 44.1 g/mol
Moles propane = 0.173 moles
Step 4: Calculate moles CO2
For 1 mol propane we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 0.173 moles propane we'll have 3*0.173 = 0.519 moles CO2
Step 5: Calculate volume of CO2
1 mol = 22.4 L
0.519 moles = 22.4 L * 0.519 = 11.6 L
11.6 L of CO2 will be produced
What evidence indicates that a reaction has occurred? (Select all that apply.)
a. The temperature decreased.
b. A solid brown product formed.
c. The temperature increased.
d. A gas formed.
e. An explosion occurred.
Final answer:
Evidence of a chemical reaction can be determined by changes such as temperature fluctuations, formation of a precipitate, gas production, and color changes. Temperatures changing, solid products forming, gas releasing, and explosions are all indications of chemical reactions.
Explanation:
Evidence indicating that a chemical reaction has occurred includes:
Temperature change: This can be an increase or a decrease, signifying an energy transfer during the reaction. Formation of a precipitate: A solid product forming in a previously clear solution is a tell-tale sign of a chemical change.Gas formation: The appearance of bubbles that are not the result of the substance boiling indicates a new gas is being produced.Color change: An unexpected change in color suggests that new substances with different properties are being formed.Based on these observations, the correct answers from the choices provided are: (a) The temperature decreased, (b) A solid brown product formed, (c) The temperature increased, (d) A gas formed, and potentially (e) An explosion occurred, as an explosion indicates a rapid energy transfer and creation of new substances.