The diagram below shows some subatomic particles. A diagram similar to a flowchart starts with a circle labeled Atom. From there, a line leads to a small circle labeled Electron and a larger one labeled Nucleus. From nucleus, two lines lead to a group of circles, labeled Neutrons and protons. From there, two lines lead to a circle surrounding two blue and one orange smaller circles. One of the blue circles is labeled with a question mark. What is the particle that is labeled with a question mark (?) in the diagram?
Answers
This question is incomplete because the options are missing. Here are the options
A. Beta particles
B. Quark
C. Isotope
D. Alpha particle
The particle labeled with a question mark in the diagram is a quark (option B)
A quark is a term used in chemistry and physics to refer to subatomic particles. These subatomic particles join together to form others such as protons and neutrons.
There are several types of quarks that are:
Quark u - upQuark d - downQuark c - charmQuark s - strangeQuark t - top / truthQuark b - bottom / beautyAccording to the above and the diagram presented, it can be inferred that the particle that is labeled with a question mark is a quark (option B)
Learn more in: https://brainly.com/question/11001639
Answer:
B
Explanation:
Related Questions
The cytochromes are heme‑containing proteins that function as electron carriers in the mitochondria. Calculate the difference in the reduction potential (ΔE∘′) and the change in the standard free energy (ΔG∘′) when the electron flow is from the carrier with the lower reduction potential to the higher. cytochrome c1 (Fe3+)+e−↽−−⇀cytochrome c1 (Fe2+)E∘′=0.22 V cytochrome c (Fe3+)+e−↽−−⇀cytochrome c (Fe2+)E∘′=0.254 V Calculate ΔE∘′ and ΔG∘′ .
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The change in reduction potential is [tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]
The change in standard free energy is [tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]
Explanation:
From the question we are told that
At the anode
[tex]cytochrome \ c_1 \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c_1 \ (Fe^{2+}) \ \ E^o = 0.22 \ V[/tex]
At the cathode
[tex]cytochrome \ c \ (Fe^{3+}) + e^-[/tex]⇔[tex]cytochrome \ c \ (Fe^{2+}) \ \ E^o = 0.254 \ V[/tex]
The difference in the reduction potential is mathematically represented as
[tex]\Delta E^o = E^o_{cathode} - E^o_{anode}[/tex]
substituting values
[tex]\Delta E^o = 0.254 - 0.220[/tex]
[tex]\Delta E^o=E^o_{cell} = 0.034 V[/tex]
The change in the standard free energy is mathematically represented as
[tex]\Delta G^o = -n * F * E^o_{cell}[/tex]
Where F is the Faraday constant with value F = 96485 C
and n i the number of the number of electron = 1
So
[tex]\Delta G^o = -(1) * 96485 * 0.034[/tex]
[tex]\Delta G^o = -3.2805 \ KJ/mol[/tex]
When ethanol, C2H5OH (a component in some gasoline mixtures) is burned in air, one molecule of ethanol combines with three oxygen molecules to form two CO2 molecules and three H2O molecules.
A) Write the balanced chemical equation for the reaction described.
B) How many molecules of CO2 and H2O would be produced when 2 molecules ethanol are consumed? Equation?
C) How many H2O molecules are formed, then 9 O2 molecules are consumed? What conversion factor did you use? Explain!
D) If 15 ethanol molecules react, how many molecules O2 must also react? What conversion factor did you use? Explain!
Answers
Answer:
1) C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)
2) four molecules of CO2 will be produced and six molecules of water
3)9 molecules of water are formed when 9 molecules of oxygen are consumed.
4) 45 molecules of oxygen
Explanation:
The balanced chemical reaction equation is shown here and must guide our work. When ethanol is burned in air, it reacts as shown;
C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)
Hence, if we use 2 molecules of ethanol, the balanced reaction equation will look like this;
2C2H5OH(l)+6O2(g)⟶4CO2(g)+6H2O(l)
Hence four molecules of CO2 are formed and six molecules of water are formed
From the balanced stoichiometric equation;
3 molecules of oxygen yields 3 molecules of water
Therefore, 9 molecules of oxygen will yield 9 × 3/3 = 9 molecules of water
Therefore, 9 molecules of water are formed when 9 molecules of oxygen are consumed.
From the reaction equation;
1 molecule of ethanol reacts with 3 molecules of oxygen
Therefore 15 molecules of ethanol will react with 15 × 3/1 = 45 molecules of oxygen
If one contraction cycle in muscle requires 55 kJ55 kJ , and the energy from the combustion of glucose is converted with an efficiency of 35%35% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.
Answers
Answer:
18 moles
Explanation:
Here the combustion of one mole of glucose ----> carbon dioxide + water, releases 2870 kilojoules / moles.
_______________________________________________________
With one contraction cycle requiring 55 kilojoules,
2870 / 55 ≈ 52.18
And with the efficiency being 35 percent,
52.1818..... * 0.35 = ( About ) 18 moles
Hope that helps!
Give one example in real life when titration can be used.
Answers
How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a final volume of 250 mL? pk, for the tris = 8.072
Answers
Answer:
41.64mL of NaOH 0.500M must be added to obtain the desire pH
Explanation:
It is possible to find pH of a buffer by using H-H equation, thus:
pH = pka + log [A⁻] / [HA]
Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.
Replacing in H-H equation:
7.60 = 8.072 + log [A⁻] / [HA]
0.3373 = [A⁻] / [HA] (1)
10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:
10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:
[A⁻] + [HA] = 0.08255 (2)
Replacing (1) in (2):
0.3373 = 0.08255 - [HA] / [HA]
0.3373[HA] = 0.08255 - [HA]
1.3373[HA] = 0.08255
[HA] = 0.06173 moles
Thus:
[A⁻] = 0.08255 - 0.06173 = 0.02082 moles [A⁻]
The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:
HA + NaOH → A⁻ + H₂O + K⁺
Thus, you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. As NaOH solution is 0.500M:
0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =
41.64mL of NaOH 0.500M must be added to obtain the desire pHAn element is a pure substance. Which of the following is used to represent an element?
Answers
Answer:
Chemists use symbols to represent elements
Explanation:
A symbol is a letter or picture used to represent something. Chemists use one or two letters to represent elements.
What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M solution of aqueous ammonia? Assume that there is no volume change upon the addition of the solid, and that the reaction goes to completion and forms Cu(NH3)42+.
Answers
Answer:
Explanation:
Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺ + 2 NO₃⁻
187.5 gm 4M 1 M
187.5 gm reacts with 4 M ammonia
18.8 g reacts with .4 M ammonia
ammonia remaining left after reaction
= .8 M - .4 M = .4 M .
187.5 gm reacts with 4 M ammonia to form 1 M Cu(NH₃)₄²⁺
18.8 g reacts with .4 M ammonia to form 0.1 M Cu(NH₃)₄²⁺
At equilibrium , the concentration of Cu²⁺ will be zero .
concentration of ammonia will be .4 M
concentration of Cu(NH₃)₄²⁺ formed will be 0.1 M
Calculate the amount of heat (kcal) released when 50.0g of steam at 100 degrees celsius hits the skin, condenses and cools to a body temperature of 37 degrees celsius
Answers
Answer:
[tex]Q=-126.1kJ[/tex]
Explanation:
Hello,
In this case, by means of the released heat, we need to consider the cooling of water in two steps:
1. Condensation of steam at 100 °C.
2. Cooling of water from 100 °C to 37 °C.
Therefore, we need the enthalpy of condensation of water that is 40.65 2258.33 J/g and the specific heat that is 4.18 J/g°C for the same amount of cooled water to obtain:
[tex]Q=50.0g*[-2258.33\frac{J}{g}+4.18\frac{J}{g\°C}(37-100)\°C]\\\\Q=-126.1kJ[/tex]
Best regards.
Apart from carbon, name other element which exhibits (slow) allotropy
Answers
Answer:
Apart from carbon, Sulfur also shows allotropy and has the following allotropes.
1) Mono clinic Sulfur
2) Rhombic Sulfur
zinc metal and hydrochloric acid react together according to the following equation 2HCl (aq)+ Zn(s) = ZnCl2(aq) +H2(g) if 5.98 g Zn reacts with excess HCl at 298 K and 0.978 atm what volume of H2 can be collected
Answers
Answer:
2.29 L
Explanation:
In this question we have to start with the chemical reaction:
[tex]2HCl_(_a_q_)~+~Zn_(_s_)~->~ZnCl_2_(_a_q_)~+~H_2_(_g_)[/tex]
The reaction is already balanced. So, if we have an excess of HCl the compound that would limit the production of [tex]H_2[/tex] would be Zn. So, we have to follow a few steps:
1) Convert from grams to moles (Using the atomic mass of Zn 65.38 g/mol).
2) Convert from moles of Zn to moles of [tex]H_2[/tex] (Using the molar mass 1 mol [tex]H_2[/tex] = 1 mol Zn).
3) Convert from mol of [tex]H_2[/tex] to volume (Using the ideal gas equation PV=nRT).
First step:
[tex]5.98~g~Zn\frac{1~mol~Zn}{65.38~g~Zn}=0.0915~mol~Zn[/tex]
Second step:
[tex]0.0915~mol~Zn\frac{1~mol~H_2}{1~mol~Zn}=0.0915~mol~H_2[/tex]
Third step:
We have to remember that R = 0.082 [tex]\frac{atm*L}{mol*K}[/tex] , so:
[tex]V=\frac{0.082\frac{atm*L}{mol*K}*0.0915~mol~H_2*298K}{0.978~atm}[/tex]
[tex]V=2.29~L[/tex]
I hope it helps!
Above 882oC, zirconium has a BCC crystal structure with a = 0.332 nm. Below this temperature, zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm. Determine the percent volume change when BCC zirconium transforms to HCP zirconium. Is that contraction or expansion?
Answers
Answer:
[tex]\mathbf{\Delta V = -0.63 \%}[/tex]
Contraction
Explanation:
From the given information;
Above 88° C
zirconium has a BCC crystal structure with a = 0.332 nm
Below this temperature
zirconium has an HCP structure with a = 0.2978 nm and c = 0.4735 nm
the volume of BCC can now be:
[tex]V_{BCC} = (a)^3[/tex]
[tex]V_{BCC} = (0.332 \ nm)^3[/tex]
[tex]V_{BCC} =0.03660 \ nm^3[/tex]
the volume of HCP can now be:
[tex]V_{HCP} = (a)^2 (c) cos \ 30^0[/tex]
[tex]V_{HCP} = (0.2978)^2 (0.4735) \ cos 30[/tex]
[tex]V_{HCP} =0.03637 \ nm^3[/tex]
Ths; the volume percent change when BCC zirconium transforms to HCP zirconium can be calculated as:
[tex]\Delta V = \dfrac{V_{HCP}-V_{BCC}}{V_{BCC}} * 100 \%[/tex]
[tex]\Delta V = \dfrac{0.03637 \ nm^3-0.03660\ nm^3}{0.03660\ nm^3}} * 100 \%[/tex]
[tex]\mathbf{\Delta V = -0.63 \%}[/tex]
Hence; it is contraction due to what the negative sign portray, The negative sign signifies that there is contraction during cooling
Choose all that apply. According to the theory of matter proposed by John Dalton: all atoms of the same element have the same mass combinations of atoms create chemical change all matter is made up of atoms all atoms of the same element have the same size
Answers
Answer: all matter is made up of atoms
all atoms of the same element have the same mass combinations of atoms create chemical change
all atoms of the same element have the same size
Explanation:
Dalton's Atomic theory suggests that all matter are made up of atoms. All atoms are made up of same elements. The atoms of the same elements will have similar physical and chemical properties. The atoms will have same size, mass, and will show similar chemical changes. The atoms in the elements are indestructible blocks and indivisible.
Where are valence electrons located in an atom?
A. In all energy levels
B. In the outermost energy level
C. In the lowest energy level
D. In the nucleus
Answers
Answer:
B.
Explanation:
Valence electrons located in the outermost energy level.
I just had that question
Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer?
Answers
Answer:
[tex]T2=276K[/tex]
Explanation:
Given:
Initial volume of the balloon V1 = 348 mL
Initial temperature of the balloon T1 = 255C
Final volume of the balloon V2 = 322 mL
Final temperature of the balloon T2 =
To calculate T1 in kelvin
T1= 25+273=298K
Based on Charles law, which states that the volume of a given mass of a ideal gas is directly proportional to the temperature provided that the pressure is constant. It can be applied using the below formula
[tex](V1/T1)=(V2/T2)[/tex]
T2=( V2*T1)/V1
T2=(322*298)/348
[tex]T2=276K[/tex]
Hence, the temperature of the freezer is 276 K
Answer: 276 kelvins
Explanation:
describe how would you use chromatography to show whether blue ink contains a single purple dye or a mixture of dyes
Answers
Explanation:
if the solution placed on the chromatography is pure there will be formation of one spot from the baseline and will go farthest to the front line unlike the impure one
With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water, and the different colors of the ink are revealed as the water moves up the paper. All of this is made possible by the water base and variety of salabilities or densities that make up ink.
What is chromatography ?Separating mixture's constituent parts by chromatography is a method. The mixture is dissolved in a material known as the mobile phase to start the process, which then transports it through a material known as the stationary phase.
A little dot of the ink to be separated is placed at one end of a strip of filter paper to perform ink chromatography. The paper strip's opposite end is submerged in a solvent. The solvent moves up the paper strip, dissolving the chemical combination as it goes and pulling it up the paper.
Throughout the experiment, the dyes are pulled along by the mobile phase (water) as it gently advances up the stationary phase (paper).
Thus, With ink chromatography, a small amount of ink is added to the paper, one end is submerged in water.
To learn more about chromatography, follow the link;
https://brainly.com/question/11960023
#SPJ5
During the winter months, many locations experience snow and ice storms. It is a common practice to treat roadways and sidewalks with salt, such as NaCl . If a 11.3 kg bag of NaCl is used to treat the sidewalk, how many moles of NaCl does this bag contain
Answers
Answer:The moles of NaCL in the 11.3kg bag is 193.36moles
Explanation:
Given that a bag of NaCl = 11.3kg
1kg = 1000g
therefore 11.3 kg = 11,300g
Remember that
No of moles = mass of subatance/ molar mass of substance
The molar mass of NaCl = Na + Cl= 22.989769 + 35.453 =58.442769≈ 58.44g/mol
No of moles = mass of subatance/ molar mass of substance
= 11300g/ 58.44g/mol = 193.36 moles
The moles of NaCL in the 11.3kg bag is 193.36 moles.
A chemist prepares a solution of barium chlorate by measuring out of barium chlorate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.
Answers
Complete Question
A chemist prepares a solution of barium chlorate BaClO32 by measuring out 42.g of barium chlorate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration is [tex]C = 0.28 \ mol/L[/tex]
Explanation:
From the question we told that
The mass of [tex]Ba(ClO_{3})_2[/tex] is [tex]m_b = 42 \ g[/tex]
The volume of the solution [tex]V_s = 500 mL = 500*10^{-3} L[/tex]
Now the number f moles of [tex]Ba(ClO_{3})_2[/tex] in the solution is mathematically represented as
[tex]n = \frac{m_b}{Z_b}[/tex]
Where [tex]Z_b[/tex] is the molar mass of [tex]Ba(ClO_{3})_2[/tex] which a constant with a value
[tex]Z_b = 304.23 \ g/mol[/tex]
Thus
[tex]n = \frac{42}{304.23}[/tex]
[tex]n = 0.14 \ mol[/tex]
The concentration of [tex]Ba(ClO_{3})_2[/tex] in the solution is mathematically evaluated as
[tex]C = \frac{n}{V_2}[/tex]
substituting values
[tex]C = \frac{0.14}{500*10^{-3}}[/tex]
[tex]C = 0.28 \ mol/L[/tex]
How many moles of each product form when the given amount of each reactant completely reacts. C3H8(g)+5O2yields 3CO2(g)+4H2O(g). 4.6 moles of C3H8
Answers
Answer: 13.8 moles of [tex]CO_2[/tex] and 18.4 moles of [tex]H_2O[/tex] will be produced
Explanation:
The given balanced reaction is;
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
Given : 4.6 moles of [tex]C_3H_8[/tex]
According to stoichiometry :
1 mole of [tex]C_3H_8[/tex] give = 3 moles of [tex]CO_2[/tex]
Thus 4.6 moles of [tex]C_3H_8[/tex] will give =[tex]\frac{3}{1}\times 4.6=13.8moles[/tex] of [tex]CO_2[/tex]
1 mole of [tex]C_3H_8[/tex] give = 4 moles of [tex]H_2O[/tex]
Thus 4.6 moles of [tex]C_3H_8[/tex] give =[tex]\frac{4}{1}\times 4.6=18.4moles[/tex] of [tex]H_2O[/tex]
Thus 13.8 moles of [tex]CO_2[/tex] and 18.4 moles of [tex]H_2O[/tex] will be produced from the given moles of reactant [tex]C_3H_8[/tex]
Draw structural formulas for all the alkene(s) formed by treatment of each haloalkane or halocycloalkane with sodium ethoxide in ethanol. Assume that elimination occurs by an E2 mechanism.
Answers
Answer:
Explanation:
Kindly note that I have attached the complete question as an attachment.
Here, we are told that elimination occurs by an E2 mechanism. What this means is that the hydrogen and the halogen must be above and below for the reaction to proceed.
The possible products are as follows;
Please check attachment for complete equations and diagrams of compounds too.
Look at the picture and observations below.
Observations: The bee's wings are moving very fast.
The bee's wings are much smaller than its body.
what’s the answer ?
Answers
Answer:
How are bees able to fly?
Explanation:
A protein's secondary structure arises due to interactions between amino acids. Two common secondary structures found in many proteins are alpha helices and beta sheets. Classify the following statements as pertaining to alpha helices or beta sheets.
1. Keep their shape due to hydrogen bonds between adjacent polypeptides chains
2. form long thin structures important for transmembrane proteins
3. flat zig zag like structure
4. a rigid spring like structure around a central axis
Answers
Answer:
Alpha helixes:
- form long thin structures important for transmembrane proteins
- have a rigid spring like structure around a central axis
Beta sheets:
- Keep their shape due to hydrogen bonds between adjacent polypeptides chains
- have a flat zig zag like structure
Explanation:
Alpha helices and beta pleated sheets are two types of secondary structure found in proteins.
Alpha helix: In this structure, the polypeptide backbone is tightly wound around an imaginary central axis drawn longitudinally through the center with the R groups of the amino acids protruding outward from the helical backbone. This structure looks like a spring and could either be a left-handed or right-handed helix, though the left-handed helix has not been observed in proteins. Each turn of the helix includes about 3.6 amino acid residues.
The alpha helix is stabilized by a hydrogen bond between the hydrogen atom attached to the electronegative nitrogen atom of a peptide linkage and the electronegative carbonyl oxygen atom of the fourth amino acid on the amino-terminal side of that peptide bond.
Alpha-helices due to their structure, are the most common transmembrane proteins- protein structure element that crosses biological membranes.
Beta sheets: In the beta conformation, the backbone of the polypeptide chain is extended into a zigzag structure. The zigzag polypeptide chain can be arranged side by side to form a structure resembling a series of pleats known as beta sheets. Hydrogen bonds formed between adjacent segments of the polypeptide chain functions to stabilize the structure.
The beta comformation in the form of turns is common in globular proteins.
a solid x when heated gives up a brown gas. If x is soluble in excess sodium hydroxide solution but insoluble in excess ammonium hydroxide solition. What is X?
Answers
Answer:
Lead (ii) nitrate
Explanation:
It is soluble in sodium hydroxide but insoluble in aqueous ammonia
When heated it produces nitrogen (IV) oxide that isbrown in colour
Under certain conditions, the substances zinc oxide and water combine to form zinc hydroxide. If 30.1 grams of zinc oxide and 6.7 grams of water combine to form zinc hydroxide, how many grams of zinc hydroxide must form
Answers
Answer:
36.8g of Zinc hydroxide
Explanation:
Based on the reaction:
ZnO + H₂O → Zn(OH)₂
Where 1 mole of zinc oxide reacts with 1 mole of water to produce 1 mole of zinc hydroxide.
Moles of 30.1g of ZnO (FW = 81.38g/mol) are:
30.1g ZnO ₓ (1mol / 81.38g) = 0.370 moles of ZnO
And moles of 6.7g of H₂O (FW = 18.01g/mol) are:
6.7g H₂O ₓ (1mol / 18.01g) = 0.372 moles of H₂O
As 1 mole of ZnO reacts per mole of H₂O, limiting reactant is ZnO because has a less number of moles than water.
Thus, moles of Zn(OH)₂ produced are 0.370 moles.
As Molar mass of Zinc hydroxide is 99.424 g/mol, there are formed:
0.370 moles Zn(OH)₂ ₓ (99.424g / mol) =
36.8g of Zinc hydroxideWhy does a chemical change occur when copper is heated?
Answers
Answer:
When copper is heated, it decomposes to form copper oxide and carbon dioxide. It is an endothermic reaction, which means that it absorbs heat. When heated, copper is easily bent or molded into shapes.
Explanation:
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compressor. Determine the compressor power per unit mass flow rate if the device is (a) isentropic, (b) polytropic with n =1.3, (c) isothermal
Answers
Answer:
(a) [tex]W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]
(b) [tex]W_{polytropic}=7.579\frac{kJ}{mol}[/tex]
(c) [tex]W_{isothermal}=5.743\frac{kJ}{mol}[/tex]
Explanation:
Hello,
(a) In this case, since entropy remains unchanged, the constant [tex]k[/tex] should be computed for air as an ideal gas by:
[tex]\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\[/tex]
[tex]0.2856=1-\frac{1}{k}\\\\k=1.4[/tex]
Next, we compute the final temperature:
[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K[/tex]
Thus, the work is computed by:
[tex]W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}[/tex]
(b) In this case, since [tex]n[/tex] is given, we compute the final temperature as well:
[tex]T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K[/tex]
And the isentropic work:
[tex]W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}[/tex]
(c) Finally, for isothermal, final temperature is not required as it could be computed as:
[tex]W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}[/tex]
Regards.
Why need to add NaAlF6 to Al2O3?
Answers
So in the electrolytic reduction of alumina, cryolite is added along with fluorspar to–
– decrease melting point of alumina
– decrease viscosity of electrolyte (CaF
2is used) – increase conductivity
Hope this helps
how many grams are there in 9.4x10^25 molecules of H2
Answers
Answer:
You start with 9.4 x 1025 molecules of H2.
You know that an Avogadro's number of molecules of H2 has a mass of 2.0 g.
To solve, 9.4 x 1025 molecules H2 x (2.0 g H2 / 6.023 x 1023 molecules H2) = 312. g H2
Explanation:
A 0.0372-m3 container is initially evacuated. Then, 4.65 g of water is placed in the container, and, after some time, all of the water evaporates. If the temperature of the water vapor is 368 K, what is its pressure
Answers
Answer:
18.3 kilopascals
Explanation:
We are given that the volume of this container is 0.0372 meters^3, that the mass of water is 4.65 grams, and that the temperature of this water vapor ( over time ) is 368 degrees Kelvins. This is a problem where the ideal gas law is an " ideal " application.
_______________________________________________________
First calculate the number of moles present in the water ( H2O ). Water has a mass of 18, so it should be that n, in the ideal gas law - PV = nRT, is equal to 4 / 18. It is the amount of the substance.
We now have enough information to solve for P in PV = nRT,
P( 0.0372 ) = 4 / 18( 8.314 )( 368 ),
P ≈ 18,276.9
Pressure ≈ 18.3 kilopascals
Hope that helps!
The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H .
Answers
Answer:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.
For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.
The acetic acid/acetate buffer system is a common buffer used in the laboratory, the equilibrium equation for the acetic acid/acetate buffer system. The formula of acetic acid is CH3CO2H -
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
An acid buffer is a solution that contains roughly the same concentrations of a weak acid and its conjugate base.
an acetate buffer contains roughly equal concentrations of acetic acid and acetate ion.Both are in chemical equilibrium with each other.The equation is:
CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺
where CH₃CO₂H - acetic acid
and, CH₃CO₂⁻ acetate ion
Thus, CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺ is the equilibrium equation for the acetic acid/acetate buffer system.
Learn more:
https://brainly.com/question/3435382
How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
A. Elements in the same group have the same number of valence
electrons.
B. The number of valence electrons increases as the atomic number
increases.
O C. The number of valence electrons is the same for all elements on
the periodic table.
D. Elements in the same period have the same number of valence
electrons.