Answer:
Etching is used to reveal the microstructure of the metal through selective chemical attack. It also removes the thin, highly deformed layer introduced during grinding and polishing. ... The specimen is etched using a reagent.
Explanation:
hope it was helpful.....
What would happen to the pressure of a closed sample of gas whose temperature increased while its volume decreased? Explain your reasoning in terms of the kinetic molecular theory of gases.
Answer:
As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.
I hope this will help you if not soo sorry :)
assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the initial mixture at equilibrium? If not, in what direction must the reactionproceed to reach equilibrium? (Hint: You will need to use the value of Kc you determined in the lab
Answer:
The mixture is not in equilibrium, the reaction will shift to the left.
Explanation:
Based on the equilibrium:
Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺
kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]
Where [] are concentrations at equilibrium. The reaction is in equilibrium when the ratio of concentrations = kc
Q is the same expression than kc but with [] that are not in equilibrium
Replacing:
Q = [10.0M] [1.0M] / [0.1M] [0.1M]
Q = 1000
As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc
carbon dioxide gas evolve during the fermentation of sugar which was collected at 22.5°C and 0.945 ATM after perfect strangers in the volume was found to be 25.0 ML how many grams of carbon dioxide were collected
Answer:
0.043 grams
Explanation:
We can find the mass of carbon dioxide as follows:
[tex] m = n*M [/tex]
Where:
n: is the number of moles
M: is the molar mass = 44.01 g/mol
First, we need to calculate the number of moles. We can use the Ideal gas equation:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 0.945 atm
V: is the volume = 25.0 mL
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the tempearture = 22.5 °C
[tex]n = \frac{PV}{RT} = \frac{0.945 atm*25 mL*\frac{1 L}{1000 mL}}{0.082 L*atm/K*mol*(22.5 + 273) K} = 9.75 \cdot 10^{-4} moles[/tex]
Hence, the mass is:
[tex]m = 9.75 \cdot 10^{-4} moles*44.01 g/mol = 0.043 g[/tex]
Therefore, 0.043 grams were collected.
I hope it helps you!
heating, the particle _______________ increases as more __________ __________ is added
Answer: what are the choices?!.
Explanation:
A solution of acetic acid that has a concentration of 0.10 moles per liter has a pH of 2.87. What is the likely pH of a 0.10 mole per liter solution of the conjugate base sodium acetate?
A. 8.97
B. 1.00
C. 2.87
D. 4.74
E. 13.00
Answer: The correct option is A) 8.97
Explanation:
To calculate the [tex]K_b[/tex] of a reaction, we use the equation:
[tex]K_a\times K_b=10^{-14}[/tex]
where,
[tex]K_a[/tex] = acid dissociation constant of acetic acid = [tex]1.86\times 10^{-5}[/tex]
[tex]K_b[/tex] = base dissociation constant
Putting values in above equation, we get:
[tex](1.86\times 10^{-5})\times K_b=10^{-14}\\\\K_b=\frac{10^{-14}}{1.86\times 10^{-5}}=5.37\times 10^{-10}[/tex]
To calculate the hydroxide ion concentration of conjugate base, we use the equation:
[tex][OH^-]=\sqrt{K_b\times \text{[Conjugate base]}}[/tex]
where,
[Conjugate base] = 0.10 M
Putting values in above equation, we get:
[tex][OH^-]=\sqrt{(5.37\times 10^{-10})\times 0.1}[/tex]
[tex][OH^-]=7.33\times 10^{-6}[/tex]
To calculate the pOH of the solution, we use the equation:
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (7.33\times 10^{-6})[/tex]
[tex]pOH=5.03[/tex]
To calculate the pH of the solution, we use the equation:
pH + pOH = 14
pH + 5.03 = 14
pH = (14 - 5.03) = 8.97
Hence, the correct option is A) 8.97
For each molecule, specify the polarity of the bonds and the overall polarity of the molecule.
a. BeCl2
b. H2O
c. O3
Dugongs are animals that live in the ocean and eat underwater grasses. The sun is shining on the shallow ocean water where the grasses and dugongs live.
What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both?
Answer:
please mark as brainliest
Explanation:
The sun is shining on the shallow ocean water where the grasses and dugongs live. What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both? Carbon is not moving into the water; it is only moving out of the water.
5 compounds that has electrovalent and covalent bond
Answer:
electrovalent
NaCl
Lithium Carbonate
ammonium phosphate
aluminium floride
potassium hydride
covalent
methane
benzene
carbon iv oxide
hydro flouride
hydro chloride
Which is a property of barium (Ba)?
O A. It rarely reacts with other elements.
O B. It is brittle as a solid.
O c. It is very reactive.
O D. It does not conduct electricity.
Plzzzzz helppppp!!!
Answer:
a it rarely reach with other elements
Which event is an example of melting?
A. Wax drips down the side of a lit candle.
B. Perspiration dries on a person's skin.
C. Rain turns to ice pellets.
D. A mirror fogs up when someone takes a hot shower.
I’m just curious tbh
Answer:
A. Wax drips down the side of a lot candle.
Explanation:
The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them
Answer:
A. Wax drips down the side of a lot candle.
Explanation:
The seagulls on the beach -
A solution of KMnO4 has an absorbance of 0.539 when measured in the colorimeter. Determine the concentration of the KMnO4 given the following data for a calibration plot.
Concentration of KMNO4 (M) Absorbance
0.0150 0.081
0.0300 0.159
0.0450 0.260
0.0600 0.334
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M
why is platinum metal preferred to other metals for the flame test
Answer:
Platinum is especially good for this because it is unreactive, and does not produce a color in the flame which will mask the presence of other metals.
Hope this answer is right!
Answer:
Hey mate, here is your answer
1. Platinum doesn't impart any color to the flame.
2. It is not oxidised under the high temperature of the flame from a bunsen burner.
3. It is almost chemically inert. Even at high temperatures, it remains unattacked by free radicals / acid radicals.
Therefore, platinum wire is crucial for a flame test. Also, a platinum wire should be thoroughly cleaned before using it for a new flame test.
A platinum wire is cleaned by dipping it into concentrated HNO3 and then placing it in the non luminous part of the bunsen flame. Otherwise, the perviously tested radicals will impart color to the flame, which may cause confusion.
Explanation:
Hope it helps you
A solution has a [H3O+] of 1 × 10−5 M. What is the [OH−] of the solution?
A) 9 M
B) 14 M
C) 1 x 10^{-9}
D) 1 x 10^{-14}
What is the quantity of
heat required to raise the
temperature of 500 g of
iron by 2°C?
The specific heat capacity
of iron is 500 J/(kg °C)
Answer:
The quantity of heat required to raise the temperature of 500 g of iron by 2°C is 500 J.
Explanation:
Calorimetry is responsible for measuring the amount of heat generated or lost in certain physical or chemical processes.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state (solid, liquid or gaseous).
Its mathematical expression is the fundamental equation of calorimetry. This is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q= ?c= 500 [tex]\frac{J}{kg*C}[/tex]m= 500 g= 0.500 kgΔT= 2 CReplacing:
Q= 500 [tex]\frac{J}{kg*C}[/tex] *0.500 kg*2 C
Solving:
Q= 500 J
The quantity of heat required to raise the temperature of 500 g of iron by 2°C is 500 J.
Lab 2: paper chromatography of organic dyes
Picture of questions below.
Answer:
The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.
Explanation:
Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.
a. Compound A and compound B are constitutional isomers with molecular formula C3H7Cl. When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination rection predominates. Propose structures A and B.
b. An unknown compound with molecular formula C6H13Cl is treated with sodium ethoxide to produce 2,3-dimethyl-2-butene as the major product. Identify the structure of the unknown compound.
Answer:
história phkfk
Explanation:
guiooupigjdytrss
The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.
a. True
b. False
Answer:
The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.
Explanation:
The given statement is true.
In chromatography silica gel is used as the predominant stationary phase.
Since silica gel is a good adsorbent.
It is a polar adsorbent.
In order to remove polar components, silica gel is used as the stationary phase.
Answer is a.true.
A molecular compound has the following empirical formula: CH2O. The molar mass of the empirical formula is g. Write your answer using 3 significant figures. If the molar mass of the molecular compound is 180.0 g/mol, write the molecular formula of the compound.
Answer:
Empirical formula has a molar mass of 30.01g/mol and molecular formula is C₆H₁₂O₆
Explanation:
Molar mass of a molecule is the sum of the molar mass of each atom. In CH2O we have:
1C = 1*12.01g/mol = 12.01g/mol
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Empirical formula of CH2O is:
12.01g/mol + 2g/mol + 16g/mol = 30.01g/mol
As the molecular compound has a molar mass of 180.0g/mol the molecular formula is:
180.0g/mol / 30.01g/mol = 6 times the empirical formula. That is:
C₆H₁₂O₆What is the concentration of s solution that contains 55 mL of alcohol per 145 mL solution?
Answer:
37.9% v/v
Explanation:
Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):
volume concentration = volume of solute / volume of solution
[tex]\% v/v = 55/145= 0.379[/tex]
phương pháp VI PHÂN ĐỒ THỊ để xác định bậc phản ứng
Answer:
mwlooy kagabi jal
64 JAHA VI PHÂN KAY
What is the pCu of the resulting solution if 20.00 mL of 0.08 M EDTA (H4Y) is added to 15.00 mL of 0.10 M CuSO4 and buffered at pH 10? The Kf’ for complex CuY2- is 2.21 x 1018
Answer:
The answer is "5.4".
Explanation:
[tex]BoH + HCL =BCL +H_2o \\\\At eq \\\\N_1V_1=N_2V_2 \\\\v_2=20 \ ml\\\\[BCL]=\frac{20 \times 0.08}{20+20}=0.04\\\\pH = \frac{1}{2} [pkw - pk_b - \log e]\\\\pk_b = 2 pH - Pkw + \Log C\\\\pK_b=5.4[/tex]
pasagot po please!!
science po ito pasagot po matino need ko po!!
Explanation:
Efficiency = (output/input)×100%
70% = output/(800 W)
which means
output = 0.70×(800 W) = 560 W
what does PH scale measure
Explanation:
pH is a measure of how acidic/basic water is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.
Which of the following aqueous solutions are good buffer systems?
a. 0.34 M calcium iodide + 0.22 M sodium iodide.
b. 0.27 M ammonia + 0.38 M ammonium nitrate.
c. 0.27 M nitric acid + 0.18 M sodium nitrate.
d. 0.18 M hydrofluoric acid + 0.14 M hydroiodic acid.
e. 0.14 M calcium hydroxide + 0.28 M calcium chloride.
Answer:
b. 0.27 M ammonia + 0.38 M ammonium nitrate.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to bear to mind the fact that buffest must be prepared by using either of the following pairs:
weak acid/conjugate base
weak base/conjugate acid
So that the pH might be set constant. In such a way, since a. shows two salts, c. a strong acid with a neutral base, d, shows two acids and e. a strong base with a neutral base, we infer the correct buffer is b. 0.27 M ammonia + 0.38 M ammonium nitrate because it has a weak base (ammonia) and its conjugate acid, ammonium.
Regards!
atomic number of element is 15 write a formula of an oxide
Answer:
Atomic Number. 15=phosphorus
Valency=3
So, Oxide=P203
Consider the following reaction at 298 K.
2 SO2(g) + O2(g) → 2 SO3(g)
An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.43 atm and 2.6 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture.
______atm.
Answer and Explanation:
The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:
[tex]Kp = \frac{P^{2}_{SO_{3} } }{P_{SO_{2}} ^{2}P_{O_{2}} }[/tex]
We have the following data:
P(SO₃) = 2.6 atm
P(O₂) = 0.43 atm
We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.
Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:
[tex]P_{SO_{2} } = \sqrt{\frac{P_{SO_{3} } ^{2} }{Kp P_{O_{2} } } } = \sqrt{\frac{(2.6 atm)^{2} }{(7 x 10^{24)}(0.43 atm) } } = 1.5 x 10^{-12} atm[/tex]
Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).
A 8.29g sample of calcium sulfide was decomposed into its constituent elements, producing 4.61g of calcium and 3.68g of sulfur. Which of the statements are consistent with the law of constant composition (definite proportions)?
a. Every sample of calcium sulfide will have 44.4% mass of calcium.
b. Every sample of calcium sulfide will have 2.86 g of calcium.
c. The mass ratio of Ca to S in every sample of calcium sulfide is 1.25.
d. The ratio of calcium to sulfur will vary based on how the sample was prepared.
e. The mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%.
Answer:
d,e
Explanation:
A student named a particular compound 2-ethyl-3-methyl-2-butene. Assuming that the student's choice actually corresponded to the correct distribution of the double bond and the substituents, what is the correct IUPAC name for this compound
Answer:
2-ethyl-3-methylbut-2-ene
Explanation:
The whole idea of IUPAC nomenclature is to devise a universally accepted system of writing the name of a compound from its structure.
According to IUPAC nomenclature, the root of the compound is the longest carbon chain. The substituents are named in alphabetical order and in such a way as to give each one the lowest number. The position of the functional group is indicated accordingly.
For the compound in question, its correct IUPAC name is 2-ethyl-3-methylbut-2-ene.
How many grams of KNO3 can dissolve in 100g of water at 20°C?
Answer:
30 grams of KNO3 can be dissolved.
Explanation:
Hello there!
In this case, since the solubility is defined as the maximum amount of solute that can be dissolved in a certain amount of solvent, usually 100 grams of water as function of the temperature, we will need to recall the graph for the solubility of KNO3 as shown on the attached file.
Thus, by identifying the curve for KNO3, we realize that at a temperature of 20 °C, the solubility is about 30 grams; which means 30 grams can be dissolved in 100 grams of water at 20 °C.
Regards!
