Answer:
pure substances and mixtures
Explanation:
pure substances are further broken down into elements and compound....... A mixture is composed of different types of atoms or molecules that are not chemically bonded.
Heterogeneous and homogeneous mixtures are the two types of mixtures. While homogeneous mixtures seem consistent throughout, heterogeneous mixtures have clearly discernible components. A solution, which can be a solid, liquid, or gas, is the most typical kind of homogenous mixture.
Explain about the types of mixtures?
Mixtures are materials made up of two or more different types of matter. Physical means can be used to separate them. Examples include a salt-water solution, a sugar-water solution, various gases, air, etc. The many components of any mixture do not come together by any sort of chemical transformation.
Solutions, suspensions, and colloids are the three categories into which mixtures can be divided based on particle size. A mixture's constituent parts maintain their unique physical characteristics
Answer:
both pure materials and mixes
Explanation:
The breakdown of pure substances into their component components and compounds continues. Different kinds of unchemically linked atoms or molecules make up a mixture.
To learn more about types of mixtures refer to:
https://brainly.com/question/24869423
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Decide which element probably has a melting point most and least similar to the melting point of cesium
The question is incomplete, so the complete question is as follows:
Decide which element probably has a melting point most and least similar to the melting point of cesium. Comparing melting point: helium thallium sodium strontium
Answer:
Most similar melting point : Sodium
Least similar melting point: Strontium
Explanation:
Melting point is defined as the temperature at which soid will melt or get converted into liquid.
Melting point of cesium is 28.44 °C and most similar melting point is sodium as its melting point is 97.72°C and least similar melting point is strontium as it has melting point 777°C.
Hence, the correct answer is:
Most similar melting point : Sodium
Least similar melting point: Strontium
Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?
Answer:
Positron emission
Explanation:
Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.
A positron is usually ejected in the process together with an anti-neutrino to balance the spins.
A scientist wants to use a model to help present the results of his detailed scientific investigation.
Why would a model be useful?
because the model makes the concepts easier to understand
because the model is easy to put together and to use
because the model prevents other scientists from asking questions
Obecause the model requires the audience to pay full attention to it
Answer:
Because the model makes the concepts easier to understand
Explanation:
Models are created to give a visual of every aspect of an experiment. This allows for a better understanding across the board for everyone.
The number of possible monobromination products, including cis-trans isomers, of methylcyclopentane is:________
A) 2
B) 3
C) 4
D) 5
E) 6
Answer:
D) 5
Explanation:
In this case, the "monobromination" is the addition of one "Br" to the molecule. With this in mind, we have to explore each option to put this "Br" atom:
1) (bromomethyl)cyclopentane = The Br atom is placed in carbon 6
2) 1-bromo-1-methylcyclopentane = The Br atom is placed in carbon 1
3) (1R,2R)-1-bromo-2-methylcyclopentane = The Br atom is placed in carbon 2 (with trans configuration)
4) (1R,2S)-1-bromo-2-methylcyclopentane = The Br atom is placed in carbon 2 (with cis configuration)
5) 1-bromo-3-methylcyclopentane = The Br atom is placed in carbon 3
See figure 1
I hope it helps!
The following compounds have similar molecular weights. Which has the highest boiling point?
A. CH3OCH3
B. C2H5OH
C. CH3CH2CH3
D. CH3CH=O
Answer:
[tex]\Huge \boxed{\mathrm{C_2H_5OH}}[/tex]
Explanation:
CH₃OCH₃ (Dimethyl ether) has a boiling point of -23 °C
C₂H₅OH (Ethanol) has a boiling point of 78.37 °C
CH₃CH₂CH₃ (Propane) has a boiling point of -42 °C
CH₃CHO (Acetaldehyde) has a boiling point of 20.2 °C
C₂H₅OH (Ethanol) has the highest boiling point.
list any five items that can be found in a Science portfolio
Answer:
physical projects
Journal entries
Materials
Lab reports
Artworks
Explanation:
Definition of a Portfolio:
Portfolio can be defined as a physical collection of student work that includes materials such as written assignments, completed tests, artwork, lab reports, physical projects and other material evidence of learning progress and academic accomplishment, including awards and honors,
A portfolio is a long-term form of self reflection and assessment that students do together.
Portfolios are a great way to demonstrate the competencies you would list on a resume or talk about in a science interview
a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P4O6 reacts further to undergo reaction B. What is the limiting reactant for the formation of P4O6? b) What mass of P4O6 is produced (theoretical yield in grams)? c) If 7.12g of P4O6 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?
The given question is incomplete, the complete question is:
Balance the following chemical equations: A) P4+ O2 → P406 B) _P406+ LO2 → P4010 a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P406 reacts further to undergo reaction B. What is the limiting reactant for the formation of P406? b) What mass of P406 is produced (theoretical yield in grams)? c) If 7.12g of P406 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?
Answer:
The balanced reaction will be,
A) P₄ + 3O₂ ⇒ P₄O₆
B) P₄O₆ + 2O₂ ⇒ P₄O₆
a) Based on the given information, the reaction container holds 5.33 grams of P₄ and 3.77 grams of oxygen. Thus, the moles of P₄ will be,
Moles = mass of P₄/Molar mass of P₄ = 5.33 grams/124 g/mole = 0.043 mole
Now the moles of O₂ will be,
Moles = mass of O₂/Molar mass of O₂ = 3.77 grams/32 g/mol = 0.112 mole
Now the moles of P₄O₆ formed when 0.043 moles of P₄ react completely will be = 1/1 × 0.043 = 0.043 mole of P₄O₆
Similarly, the moles of P₄O₆ formed, when 0.112 moles of O₂ react completely will be = 1/3 × 0.112 = 0.0373 mole of P₄O₆
Thus, from the analysis, the maximum moles of P₄O₆ formed will be 0.0373 moles. Therefore, oxygen will be the limiting reagent, which will react completely in the reaction.
b) From the above findings, the maximum moles of P₄O₆ produced is 0.0373 mole. Thus, the theoretical yield of P₄O₆ produced will be,
= Moles of P₄O₆ × Molar mass of P₄O₆
Theoretical yield = 0.0373 mole × 220 g/mole = 8.206 grams
c) Based on the given information, the actual mass of P₄O₆ produced is 7.12 grams.
Hence, percent yield = Actual yield/Theoretical yield * 100
= 7.12/8.206 × 100 = 86.77 %
d) In the given case, reaction B will not take place. This is due to the fact that oxygen is not left for reaction B, which was the limiting regent for reaction A. Here P₄ is the excess reactant, which was left in the reaction.
The initial moles of P₄ is 0.043, O₂ is 0.112, and P₄O₆ is O. The final moles of P₄ is 0.043 -1/3 × 0.112 = 0.0057 mole, O₂ is 0, and P₄O₆ is 0.0373 mole.
Thus, moles of P₄ left is 0.0057 mole. Hence, the mass of P₄ left will be,
= 0.0057 mole × Molar mass of P₄
= 0.0057 mole × 124 g/mole = 0.7068 grams.
A sheet of aluminum foil weighs 2.07g. If the sheet is 24cm long and 20cm wide, how thick is the sheet in micrometers? The density of aluminum is 2.7 g/cm3.
Answer:
[tex]h=1.60\mu m[/tex]
Explanation:
Hello,
In this case, given the density and the mass of the aluminum foil, we can compute the occupied volume as shown below:
[tex]\rho =\frac{m}{V}\\ \\V=\frac{m}{\rho}=\frac{2.07g}{2.7g/cm^3} =0.767cm^3[/tex]
Next, since the volume is defined as:
[tex]V=24cm*20cm*h[/tex]
Whereas [tex]h[/tex] accounts for its thickness, we can find it to be:
[tex]h=\frac{V}{24cm*20cm}=\frac{0.767cm^3}{20cm*24cm}\\ \\h=1.60x10^{-3}cm*\frac{10000\mu m}{1cm} \\\\h=1.60\mu m[/tex]
Regards.
A buffer solution is 0.413 M in HF and 0.237 M in KF. If Ka for HF is 7.2×10-4, what is the pH of this buffer solution?
Answer:
2.90
Explanation:
Any buffer system can be described with the reaction:
[tex]HA~->~H^+~+~A^-[/tex]
Where [tex]HA[/tex] is the acid and [tex]A^-[/tex] is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:
[tex]pH=pKa~+~Log\frac{[A^-]}{[HA]}[/tex]
With all this in mind, we can write the reaction for our buffer system:
[tex]HF~->~H^+~+~F^-[/tex]
In this case, the acid is [tex]HF[/tex] with a concentration of 0.413 M and the base is [tex]F^-[/tex] with a concentration of 0.237 M. We can calculate the pKa value if we do the "-Log Ka", so:
[tex]pKa~=~-Log(7.2X10^-^4)=~3.14[/tex]
Now, we can plug the values into the Henderson-Hasselbach
[tex]pH=~3.14~+~Log(\frac{[0.237~M]}{[0.413~M]})~=~2.90[/tex]
The pH value would be 2.90
I hope it helps!
Calculate the Ka for the following acid. Determine if it is a strong or weak acid.
HNO2(aq) dissolves in aqueous solution to form H+(aq) and NO2−(aq). At equilibrium, the concentrations of each of the
species are as follows:
[HNO2]=0.68M
[H+]=0.022M
[NO2−]=0.022M
Calculate the for the following acid. Determine if it is a strong or weak acid.
dissolves in aqueous solution to form and . At equilibrium, the concentrations of each of the species are as follows:
a) Ka=7.1×10−4; This is a weak acid because the acid is not completely dissociated in solution.
b) Ka=1405; This is a strong acid because the Ka is very large.
c) Ka=1405; This is a weak acid because the acid is not completely dissociated in solution.
d) Ka=7.1×10−4; This is a strong acid because the acid is completely dissociated in aqueous solution.
Answer:
a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.
Explanation:
Step 1: Write the dissociation reaction for nitrous acid
HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)
Step 2: Calculate the acid dissociation constant
Ka = [H⁺] × [NO₂⁻] / [HNO₂]
Ka = 0.022 × 0.022 / 0.68
Ka = 7.1 × 10⁻⁴
Step 3: Determine the strength of the acid
Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.
How many milliliters of a 1M nitric acid solution are required to prepare 60mL of 6.7M solution?
A) 400 mL
B) 4mL
C) 0.25 mL
D) none of the above
Answer:
the number of milliliters of a 1M is 402mL
Explanation:
The computation of the number of milliliters could be determined by using the following formula
As we know that
[tex]V_1\times M_1 = V_2\times M_2[/tex]
where,
V_1 and V_2 are the starting and final volumes
And, the M_1 and M_2 are the starting and the final molarities
Now the V_1 is
[tex]V_1 \times 1M = 60mL \times 6.7M[/tex]
So, the V_1 is 402mL
Hence, the number of milliliters of a 1M is 402mL
Fructose‑2,6‑bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose‑1,6‑bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose‑2,6‑bisphosphate is regulated by many hormones, second messengers, and enzymes.
Answer:
Activate glycolysis/Inhibit gluconeogenesis: Increased levels of fructose-2,6-bisphosphate, activation of PFK-2
Activate gluconeogenesis/Inhibit glycolysis: Increased levels of glucagon, Increased levels of cAMP, Activation of fructose-2,6-bisphosphatase (FBPase-2)
Note: The question is incomplete. The complete question is given below and in the attachment.
Fructose‑2,6‑bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose‑1,6‑bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose‑2,6‑bisphosphate is regulated by many hormones, second messengers, and enzymes. How do the following affect glycolysis and gluconeogenesis?
Explanation:
Fructose-2,6-bisphosphate is an allosteric effector for the enzymes phosphofructokinase-1 (PFK-1) and fructose-1,6-bisphosphatase (FBPase-1). It increases the affinity of PFK-1 for fructose-6-phosphate thereby activating glycolysis. However, it reduces the affinity of FBPase-1 for its substrate, fructose-1,6-bisphosphate thereby inhibiting gluconeogenesis.
Activation of phosphofructokinase-2 activates glycolysis and inhibits gluconeogenesis by catalyzing the phosphorylation of fructose-6-phosphate to form fructose-2,6-bisphosphate.
Increased levels of glucagon stimulates the synthesis of cAMpP which activates cAMP-dependent ptrotein kinase which phosphorylates the bifunctional enzyme PFK-2/FBPase-2. The phosphorylation of this enzyme inhibits its PFK-2 activity and activates its FBPase-2 activity. This results in the activation of gluconeogenesis and inhibition of glycolysis.
Fructose-2,6-bisphosphatase breaks down fructose-2,6-bisphosphate to fructose-6-phospshate and a phosphoryl group. This results in the activation of gluconeogenesis and the inhibition of glycolysis.
The following equilibrium is formed when copper and bromide ions are placed in a solution:
heat + Cu(H2O)6 ^+2 (blue) + 4Br- <--> 6H2O + CuBr4^-2 (green)
A) answer the following questions when KBr is added to the solution:
1. What will happen to the equilibrium?
2. What will be the color of the solution?
3. Will the solution be hotter or cooler? Explain.
B) What will be the color of the solution when the solution is heated?
Answer:
A)
1. Reaction will shift rightwards towards the products.
2. It will turn green.
3. The solution will be cooler..
B) It will turn green.
Explanation:
Hello,
In this case, for the stated equilibrium:
[tex]heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)[/tex]
In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:
A)
1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.
2. The formation of the green complex is favored, therefore, it will turn green.
3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.
B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.
Regards.
What type of matter is pepperoni pizza
Answer:
Heterogeneous Mixture. Have a good day! =)
Explanation:
What is the density of an object with a mass of 145.8g and an volume of 91.75 mL?
Answer:
Density = 1.6 g/mLExplanation:
Density of a substance can be found by using the formula
[tex]Density = \frac{mass}{volume} [/tex]
From the question
mass = 145.8 g
volume = 91.75 mL
Substitute the values into the above formula and solve for the Density
That's
[tex]Density = \frac{145.8}{91.75} [/tex]
= 1.5891
We have the final answer as
Density = 1.6 g/mL
Hope this helps you
Which of the following definitions best describes the term "vapor pressure?a. Pressure and temperature values on a phase diagram where two phases of a substance coexist. b. In equilibrium with the liquid phase, the pressure exerted by a gas. c. Aspecific temperature and pressure at which the liquid and gas phases of a substance have the same density and are indistinguishable from each other. d. The temperature and pressure at which all three phases of a substance coexist. Under these conditions, freezing and melting, boiling and liquefaction, and sublimation and deposition all proceed at the same rate. e. The temperature at which the vapor pressure of a liquid equals 1 atm.
Answer:
b. In equilibrium with the liquid phase, the pressure exerted by a gas.
Explanation:
When a liquid is warmed up to a temperature , it starts vaporising . The liquid is turning into gas and gas is turning into liquid at different rates . Initially the rate of former is higher but gradually the difference of rate between them decreases to zero . At this point the rate of conversion of liquid into gas and rate of conversion of gas into liquid becomes equal . This is called dynamic equilibrium point .
If we change the temperature , the equilibrium gets disturbed .
At this point the pressure exerted by the gas is called the vapour pressure of the liquid .
So option b ) is correct .
At the equilibrium point, the pressure exerted by the gas molecules to the liquid molecules has been termed the Vapor pressure. Thus, statement b is correct.
The vaporization has been the process of conversion of liquid to the gaseous state with the rise in temperature. The liquids attaining a certain temperature have been vaporized into the gaseous state.
Initially, the gas phase has been less in concentration, thus the rate of formation of gas has been greater.
After a certain amount of time, the gas phase starts to cool down and converts to the liquid state. The rate of formation of the liquid has been slower.
The time when the rate of formation of liquid, and the rate of formation of gas has been equal is termed as the equilibrium point. At the equilibrium point, the pressure exerted by the gas molecules to the liquid molecules has been termed the Vapor pressure.
Thus, statement b is correct.
For more information about the Vapor pressure, refer to the link:
https://brainly.com/question/8646601
Which one of the following statements a about scientific hypothesis is FALSE? A scientific hypothesis is an educated guess about why something happens. In order to be useful, a scientific hypothesis must be testable in a way that is replicable by other scientists. The previously known outcome of an observation or experiment can be used as solid proof that a newly-created scientific hypothesis is absolutely true. A scientific hypothesis is an explanation for a natural phenomenon.
Answer:
C). The previously known outcome of an observation or experiment can be used as solid proof that a newly-created scientific hypothesis is absolutely true.
Explanation:
A Scientific hypothesis is characterized as the proposed explanation or an educated guess about a natural phenomenon on the basis of previous knowledge as well as observation. All the given statements are true regarding a scientific hypothesis that it is 'an educated guess which explains the reason why a specific phenomenon occurs', 'being testable in a manner that could be replicated by other', 'an explanation for a natural phenomenon' except for that the 'truth of a scientific hypothesis can never be assured completely with a solid proof as it always has chances of being expanded.' Thus, option C is the correct answer.
Calculate the solubility of BaCO3 (a) in pure water and (b) in a solution in which [CO32-] = 0.289 M. Solubility in pure water = M Solubility in 0.289 M CO32- = M
Answer:
the solubility of BaCO₃ in pure water and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.
Explanation:
To calculate the solubility of BaCO₃ in:
(a) pure water and (b) in a solution in which [CO₃²⁻] = 0.289 M
The [tex]ksp[/tex] (i.e solubility constant ) for BaCO₃= 2.0 × 10⁻⁹
BaCO₃ → Ba²⁺ + CO₃²⁻
ksp = s × s
s² = ksp
s = [tex]\sqrt{ksp}[/tex]
s = [tex]\sqrt{2.0 \times 10^{-9}}[/tex]
s = 4.472 × 10⁻⁵ M
(b) The solubility of BaCO₃ in a solution in which [CO₃²⁻] = 0.289 M
BaCO₃ → Ba²⁺ + CO₃²⁻
ksp = s × s
2.0 × 10⁻⁹ = s × 0.289
s = 2.0 × 10⁻⁹/0.289
s = 6.9204 × 10⁻⁹ M
Thus, the solubility of BaCO₃ in pure water and in a solution is 4.472 × 10⁻⁵ M and 6.9204 × 10⁻⁹ M respectively.
What is the density of an object with a mass of 83 g and a volume of 34 mL?
Type your answer with at least 2 decimal places.
What information about earthquakes do scientists gain from seismographs?
Answer:
how strong it is
Explanation:
Seismographs are not able to say when earthquake will happen, but they help humans to know how strong it is or if it is happening or not
Consider the reaction of 1-butanol with K2Cr2O7, H2SO4, heat. Draw only the organic product derived from 1-butanol.
Answer:
Butanoic acid.
Explanation:
Hello,
In this case, when a primary alcohol such as 1-butanol (OH is bonded to a primary carbon) is oxidized in the presence of a strong oxidizing media such as potassium dichromate (K2Cr2O7) and sulfuric acid, the stepwise oxidation goes to the corresponding aldehyde with a further oxidation to the corresponding carboxylic acid:
[tex]R-CH_2-OH\longrightarrow R-COH\longrightarrow R-COOH[/tex]
Therefore, on the attached picture you can find that the formed aldehyde is butanal and the inly organic product, due to the strong oxidizing media is finally butanoic acid.
Best regards.
Al2O3 (s) + 6 NaOH (aq) + 12HF (g) → 2 Na3AlF6 (s) + 9 H2O (l) In an experiment; 6.55 g Al2O3 and excess HF were dissolved in 1.75 L of 0.15 M NaOH. If 20 g Na3AlF6was obtained, a Which one is the limiting reagent? b What is the actual yield? c What is the theoretical yield? d What is the percent yield for this experiment?
Answer:
A. NaOH
B. 20 g
C. 18.4 g
D. 108%
Explanation:
Al₂O₃ + 6 NaOH + 12 HF ⇒ 2 Na₃AlF₆ + 9 H₂O
A. Since you have excess HF, this is not the limiting reagent. The two possibilities are Al₂O₃ and NaOH. Find out how many moles you have of each. Then, use the mole ratio in the chemical equation to find out how much product each reagent can produce. The reagent the produces the least is the limiting reagent. The molar mass of Al₂O₃ is 101.96 g/mol.
(6.55 g Al₂O₃)/(101.96 g/mol Al₂O₃) = 0.06424 mol Al₂O₃
(0.06424 mol Al₂O₃) × (2 mol Na₃AlF₆/1 mol Al₂O₃) = 0.12848 mol Na₃AlF₆
(1.75 L NaOH) × (0.15 M NaOH) = 0.2625 mol NaOH
(0.2625 mol NaOH) × (2 mol Na₃AlF₆/6 mol NaOH) = 0.0875 mol Na₃AlF₆
NaOH produces less product, making it the limiting reagent.
B. The actual yield is 20 g. This information is given in the problem.
C. Since you know how much product you will get, convert moles of Na₃AlF₆ to grams to find actual yield. Use the value found for Na₃AlF₆ that you got from the limiting reagent. The molar mass is 209.94 g/mol.
(0.0875 mol Na₃AlF₆) × (209.94 g/mol Na₃AlF₆) = 18.4 g Na₃AlF₆
D. To find the percent yield, divide the actual yield by the theoretical yield and multiply by 100%.
(20 g)/(18.4 g) × 100% = 108%
If your percent yield is greater than 100% when performing a reaction, there has been a mistake somewhere. Either you recorded the numbers incorrectly or there has been some human error in your experiment. This might be the right answer for this problem, but you might want to double check to make sure the numbers you gave me were right.
how many moles of MgO are produced when .250 mol of Mg reacts completely with O2
Answer:
0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂
Explanation:
In first place, the balanced reaction between Mg and O₂ is:
2 Mg + O₂ ⇒ 2 MgO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:
Mg: 2 molesO₂: 1 moleMgO: 2 molesThen you can apply the following rule of three: if by reaction stoichiometry 2 moles of Mg produce 2 moles of MgO, 0.250 moles of Mg, how many moles of MgO will they form?
[tex]moles of MgO=\frac{0.250 moles of Mg*2 moles of MgO}{2 moles of Mg}[/tex]
moles of MgO= 0.250
0.250 moles of MgO are produced when 0.250 mol of Mg reacts completely with O₂
hen solid NH4NO3 is dissolved in water, the temperature of the water and beaker gets noticeably colder. The formation of an aqueous solution of ammonium nitrate is __________
Answer:
The formation of an aqueous solution of ammonium nitrate is An endothermic process
Explanation:
An exothermic process produce energy when occurs. As there is energy that is released, the temperature of the arounds increases.
In the other hand, an endothermic process absorb energy when occurs doing the temperature of the around colder than the initial temperature.
As the dissolution of NH₄NO₃ in water make the temperature of the water colder:
The formation of an aqueous solution of ammonium nitrate is An endothermic processThe formation of an aqueous solution of ammonium nitrate is an endothermic process
The question requires us to determine if reaction process is an endothermic or an exothermic reaction.
To do this,
First we will define the terms Endothermic reaction and exothermic reaction
Endothermic reactions are chemical reactions in which the reactants absorb heat energy from the surroundings to form products. These reactions lower the temperature of their surrounding area, thereby creating a cooling effect. They have a net positive standard enthalpy change.
Exothermic reactions are reactions or processes that release energy, usually in the form of heat or light. They have a net negative standard enthalpy change.
From the question,
When solid NH4NO3 is dissolved in water, the temperature of the water and beaker gets noticeably colder. This means it is an endothermic reaction.
Hence, the formation of an aqueous solution of ammonium nitrate is an endothermic process
Learn more here: https://brainly.com/question/2487822
The quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dissolved in hot, concentrated acid and passed over a reducing agent so that all the antimony is in the form Sb3 (aq). The Sb3 (aq) is completely oxidized by an aqueous solution of BrO3–(aq). Complete and balance the equation for this reaction in acidic solution.
6H+ +Bro3- +3Sb3+----------> Br- +3Sb5+3H2O
Answer:
3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)----->3Sb^5+(aq) + Br^-(aq) 3H2O(l)
Explanation:
When we want to balance redox reaction equations, we must ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation.
After we have done this, we can now write the overall balanced reaction equation without including the number of electrons lost or gained. Hence;
Oxidation half equation;
3Sb^3+(aq) -----> 3Sb^5+(aq) +6e
Reduction half equation;
BrO3^-(aq) + 6H^+(aq) + 6e ----> Br^-(aq) 3H2O(l)
Overall balanced reaction equation;
3Sb^3+(aq) + BrO3^-(aq) + 6H^+(aq)----->3Sb^5+(aq) + Br^-(aq) 3H2O(l)
Calculate the pZn of a solution prepared by mixing 25.0 mL of 0.0100 M EDTA with 50.0 mL of 0.00500 M Zn2 . Assume that both the Zn2 and EDTA solutions are buffered with 0.100 M NH3 and 0.176 M NH4Cl.
Answer:
[tex]\mathbf{pZn ^{2+} =8.8569 }[/tex]
Explanation:
Using the approach of Henderson-HasselBalch equation, we have :
[tex]pH = pKa[NH^+_4] + log \dfrac{[NH_3]}{[NH_4^+]}[/tex]
where;
the pKa of [tex]NH^+_4[/tex] = 9.26
concentration of [tex]NH_3[/tex] = 0.100 M
concentration of [tex]NH_4Cl[/tex] = 0.176 M
∴
the pH of the buffered solution is :
[tex]pH = 9.26 + log \dfrac{[0.100]}{[0.176]}[/tex]
[tex]pH = 9.26 + log (0.5682)[/tex]
[tex]pH = 9.26 +(-0.2455)[/tex]
[tex]pH =9.02[/tex]
The Chemical equation for the reaction of [tex]Zn ^{2+}[/tex] and EDTA is :
[tex]Zn^{2+}_{(aq)} + Y^{4-}_{(aq)} \iff ZnY^{2-} _{(aq)}[/tex]
Here;
[tex]Y^{4-}_{(aq)}[/tex] denotes the fully deprotonated form of the EDTA
The formation constant [tex]K_f[/tex] of the equation for the reaction can be represented as:
[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}[/tex] ----- (1)
The logarithm of the formation constant of Zn - EDTA complex = 16.5
[tex]K_f[/tex] = [tex]10^{16.5}[/tex]
[tex]K_f[/tex] = [tex]3.16 \times 10^{16}[/tex]
Since the formation constant in the above equation signifies that the EDTA is present in [tex]Y^{4-}[/tex],
Then:
[tex]\alpha _{Y^{4-} }= \dfrac{Y^{4-}}{C_{EDTA}}[/tex]
[tex]{Y^{4-}}= \alpha_ {Y^{4-}} \times {C_{EDTA}}[/tex]
From (1)
[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ][Y^{4-}]}[/tex]
[tex]K_f = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ \ \alpha_ {Y^{4-}} \times {C_{EDTA}}}[/tex]
∴
[tex]K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]
where;
[tex]K_f'[/tex] = conditional formation constant
[tex]\alpha _Y{^4-}[/tex] = the fraction of EDTA that exit in the form of the presences of the 4 charges .
So at equivalence point :
all the [tex]Zn^{2+}[/tex] initially in titrand is now present in [tex]ZnY^{2-}[/tex]
[tex]K_f' = K_f \times \alpha _Y{^4-}[/tex]
Obtaining the data for the value of [tex]\alpha _Y{^4-}[/tex] at the reference table:
[tex]\alpha _Y{^4-}[/tex] = [tex]5.4 \times 10^{-12}[/tex]
∴
[tex]K_f' = 3.16 \times 10^{16} \times 5.4 \times 10^{-2}[/tex]
[tex]K_f' = 1.7064 \times 10^{15}[/tex]
To calculate the moles of EDTA ,[tex]Zn^{2+}[/tex] , [tex]ZnY^{2-}[/tex] ; we have:
moles of EDTA = 0.0100 M × 0.025 L
moles of EDTA = [tex]2.5 \times 10^{-4} \ mole[/tex]
moles of [tex]Zn^{2+}[/tex] = 0.00500 M × 0.050 L
moles of [tex]Zn^{2+}[/tex] = [tex]2.5 \times 10^{-4} \ mole[/tex]
moles of [tex]ZnY^{2-}[/tex] = [tex]\dfrac{initial \ mole}{total \ volume}[/tex]
moles of [tex]ZnY^{2-}[/tex] = [tex]\dfrac{2.5 \times 10^{-4}}{ 0.025 + 0.050 }[/tex]
moles of [tex]ZnY^{2-}[/tex] = [tex]\dfrac{2.5 \times 10^{-4}}{ 0.075 }[/tex]
moles of [tex]ZnY^{2-}[/tex] = 0.0033333 M
Recall that:
[tex]K_f' = K_f \times \alpha _Y{^4-} = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]
[tex]K_f' = \dfrac{[ZnY^{2-}]}{[Zn^{2+} ] \ C_{EDTA} }[/tex]
Assume Q² is the amount of complex dissociated in [tex]ZnY^{2-}[/tex]
[tex]ZnY^{2-} \iff Zn^{2+} + C_{EDTA}[/tex]
i.e [tex]Q^2 = Zn^{2+} + C_{EDTA}[/tex]
[tex]1.707 \times 10^{15}= \dfrac{0.0033333}{Q}[/tex]
[tex]Q= \dfrac{0.0033333}{1.707 \times 10^{15}}[/tex]
[tex]Q^2= \dfrac{0.0033333}{1.707 \times 10^{15}}[/tex]
[tex]Q^2= 1.9527 \times 10^{-18}[/tex]
[tex]Q= \sqrt{1.9527 \times 10^{-18}}[/tex]
Q = [tex]1.397 \times 10^{-9}[/tex] M
[tex][Zn^{2+}]= 1.39 \times 10^{-9} \ M[/tex]
∴
[tex]pZn ^{2+} =- log [Zn^{2+}][/tex]
[tex]pZn ^{2+} = - log (1.39 \times 10^{-9} ) \ M[/tex]
[tex]\mathbf{pZn ^{2+} =8.8569 }[/tex]
What is the final pH of a solution obtained by mixing 300 ml of 0.4 M NH3 with 175 ml of 0.3 M HCl? (Kb = 1.8 x 10-5) Show all of your math steps. Do not leave us guessing as to how you got your final answer.
Answer:
pH of the final solution = 9.15
Explanation:
Equation of the reaction: HCl + NH₃ ----> NH₄Cl
Number of moles of NH₃ = molarity * volume (L)
= 0.4 M * (300/1000) * 1 L = 0.12 moles
Number of moles of HCl = molarity * volume (L)
= 0.3 M * (175/1000) * 1 L = 0.0525 moles
Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced
Number moles of NH₄Cl produced = 0.0525 moles
Number of moles of base left unreacted = 0.12 - 0.0525 = 0.0675
pOH = pKb + log([salt]/[base])
pKb = -logKb
pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)
pOh = 4.744 + 0.109
pOH = 4.853
pH = 14 - pOH
pH = 14 - 4.853
pH = 9.15
Therefore, pH of the final solution = 9.15
The percent by mass of methanol (MM = 32.04 g/mol) in an aqueous solution is 21.1%. What is the molality of the methanol solution?
Answer:
Molality of the methanol solution = 8.33 m
Explanation:
Given:
Mass % = 21.1 %
Molar mass of methanol = 32.04 g / mol
Find:
Molality of the methanol solution?
Computation:
Moles of methanol = Mass / Molar mass
Moles of methanol = = 21.1 / 32.04
Moles of methanol = 0.658
Assume.
Mass of solution = 100 g
Mass of solvent = 100 -21.1 = 78.9 g = 0.0789 kg
Molality of the methanol solution = 0.658 / 0.0789
Molality of the methanol solution = 8.33 m
Given values:
Mass percentage = 21.1%Methanol's molar mass = 32.04 g/molNow,
Moles of methanol:
= [tex]\frac{Mass}{Molar \ mass}[/tex]
= [tex]\frac{21.1}{31.04}[/tex]
= [tex]0.658[/tex]
then,
Mass of solution:
= [tex]100-21.1[/tex]
= [tex]78.9 \ g \ or \ 0.0789 \ kg[/tex]
hence,
The molality will be:
= [tex]\frac{0.658}{0.0789}[/tex]
= [tex]8.33 \ m[/tex]
Thus the answer above is correct.
Learn more about molality here:
https://brainly.com/question/17218475
which of the following changes are chemical changes?
A water if frozen
B water is heated up
C gasoline is burned
D water is boiled
E gasoline is evaporated
Answer:
the answer is C.
Explanation:
this is because burning anything is going to change the make-up of the object
40.002 g : 13.000005 g =