Answer:
A, E, and F!
Explanation:
Evidence that was gathered by scientists through the use of Fossils among the given options are;
The fossil record can be used to date rocks.
Scientists can figure out what ancient living things ate.
Pollen and seeds help provide clues about ancient climates
Fossils can be regarded as remnants or traces that was left behind by things, organisms, animals that has once a living thing from the past. Examples of Fossils could be bones as well as shells and exoskeletons or stone, imprints of animals.Fossil record can be explained as compilation both the known and unknown fossils that exist over a period of time.Some evidence gathered by scientists through the use of Fossils helps to know that with fossil record the date rocks can be known and figuring out of foods that was eaten by ancient living things. With the Pollen and seeds, clues about ancient climates can be gottenWith the study of Fossils records, the scientist can tell the period of time that life has existed here on Earth. It also tells the relationship between plant and animals.
Therefore, options A F and E are required options.
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The reaction of 15 moles carbon with 30 moles O2 will
result in a theoretical yield of __ moles CO2.
Answer:
15 moles.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]C+O_2\rightarrow CO_2[/tex]
Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:
[tex]n_{CO_2}=15molC*\frac{1molCO_2}{1molC} \\\\n_{CO_2}=15molCO_2[/tex]
Best regards.
State the effect of anion hydrolysis on the pH of water
Answer:
Depending on the anions and cations present within a hydrolysis reaction, the solution can be more... ... This lesson will explain how this occurs. ... that could react with water and create products that affect the characteristics of the solution.
Answer:
Salts of weak bases and strong acids do hydrolyze, which gives it a pH less than 7. This is due to the fact that the anion will become a spectator ion and fail to attract the H+, while the cation from the weak base will donate a proton to the water forming a hydronium ion.
Explanation:
I hope this is the answer your looking for
Duncan knows that it takes 36400 cal of energy to heat a pint of water from room temperature to boiling. However, Duncan has prepared ramen noodles so many times he does not need to measure the water carefully. If he happens to heat 0.800 pint of room-temperature water, how many kilojoules of heat energy will have been absorbed by the water at the moment it begins to boil?
Answer:
[tex]\large \boxed{\text{122 000 J}}[/tex]
Explanation:
1. Calculate the energy needed
[tex]\text{Energy} = \text{0.800 pt} \times \dfrac{\text{36 400 cal}}{\text{1 pt}} = \text{ 29 120 cal}[/tex]
2. Convert calories to joules
[tex]\text{Energy} = \text{29 120 cal} \times \dfrac{\text{4.184 J}}{\text{1 cal}} = \textbf{122 000 J}\\\\\text{The water will have absorbed $\large \boxed{\textbf{122 000 J}}$}[/tex]
Carbon dioxide and water vapor are variable gases because _____.
Answer: their amounts vary throughout the atmosphere
Explanation:
There is very little that travels over the atmosphere
Vary=very little
Answer:
Carbon dioxide and water vapor are variable gases because their amounts vary throughout the atmosphere.
Variable gases are not constant gases. Constant gases have almost uniform concentrations through out the Earth's atmosphere.
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
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 pHs the following nuclear equation balanced? yes no
Answer:
Yes.
Explanation:
The nuclear equation {226/88 Ra → 222/26 Rn + 4/2 He} is balanced. As we know that an alpha particle is identical to a helium atom. This implies that if an alpha particle is eliminated from an atom's nucleus, an atomic number of 2 and a mass number of 4 is lost.
Therefore, the equation will be reduced to:
226 - 4 = 222
88 - 2 = 86
Hence, the equation is balanced.
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.
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for water.) Determine the freezing point and boiling point of the solution. (Assume a density of 1.00 g/ mL for water.)
Answer: The freezing point and boiling point of the solution are [tex]-6.6^0C[/tex] and [tex]101.8^0C[/tex] respectively.
Explanation:
Depression in freezing point:
[tex]T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_f[/tex] = freezing point of solution = ?
[tex]T^o_f[/tex] = freezing point of water = [tex]0^0C[/tex]
[tex]k_f[/tex] = freezing point constant of water = [tex]1.86^0C/m[/tex]
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g
[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]
[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
[tex](0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]
[tex]T_f=-6.6^0C[/tex]
Therefore,the freezing point of the solution is [tex]-6.6^0C[/tex]
Elevation in boiling point :
[tex]T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of water = [tex]100^0C[/tex]
[tex]k_b[/tex] = boiling point constant of water = [tex]0.52^0C/m[/tex]
i = vant hoff factor = 1 ( for non electrolytes)
m = molality
[tex]w_2[/tex] = mass of solute (ethylene glycol) = 21.4 g
[tex]w_1[/tex]= mass of solvent (water) = [tex]density\times volume=1.00g/ml\times 97.6ml=97.6g[/tex]
[tex]M_2[/tex] = molar mass of solute (ethylene glycol) = 62g/mol
Now put all the given values in the above formula, we get:
[tex](T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}[/tex]
[tex]T_b=101.8^0C[/tex]
Thus the boiling point of the solution is [tex]101.8^0C[/tex]
How does the number of valence electrons in an atom relate to the element's
placement on the periodic table?
O 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
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.
Answer:
A
Explanation:
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!
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
The bromine test (part d) is often used as an indication of unsaturation(double and triple bonds). Explain why your result for trichloroethylene and toluene were different than for the simple alkene produc
Answer:
Toluene is an aromatic compound not an alkene
Bromine test is used to determine the presence of unsaturation in the given compound. The trichloroethylene does not have any unsaturation while toluene have double bonds of benzene ring. Therefore, the Bromine test can differentiate between trichloroethylene and toluene.
What is degree of unsaturation?The degree of unsaturation of an organic compounds can be categorised two types: saturated and unsaturated. Saturated compounds are those that have only single bonds. An unsaturated compound are those that has a double bond, triple bond, and/or ring(s).
The alkanes with only single bonds are classified as saturated whereas the alkenes and alkynes with double and triple bonds are classified as unsaturated hydrocarbons.
The degree of unsaturation formula helps in finding whether a compound is saturated or unsaturated.
In the Bromine test when the bromine solution will be added into the compound if the brown color of the solution will disappear it means the unsaturation is present in the given compound.
Therefore, the we can distinguish between trichloroethylene and toluene with bromine test.
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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∘′ .
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]
An element is a pure substance. Which of the following is used to represent an element?
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.
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
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 hydroxideThe iceman known as Otzi was discovered on a mountain on the Austrian-Italian border. Samples of his hair and bones had carbon-14 activity that was about 12.5% of that present in new hair or bone. How long ago did Otzi live if the half-life for C-14 is 5730 years
Answer:
1432.5 years
Explanation:
The rate of decay of a radioactive isotope is the characteristics of the isotope and it is usually expressed in terms of its half-life.
The half-life of a radioactive element is the time taken for half of the total number of atoms in a given sample of the element to decay or the time taken for the intensity of radiation to fall to half of its original value.
From the given question.
Since the same of his bones had a carbon-14 activity that was about 12.5% of that present in new hair or bone.
Thus; the time taken to reduce the amount of the sample to one-quarter of its amount(12.5%) = the half life for C-14 (5730 years)
The time taken for how long Otzi live = 5730/4 = 1432.5 years
Why need to add NaAlF6 to Al2O3?
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 ?
Answer:
How are bees able to fly?
Explanation:
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
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]
Consider the insoluble compound zinc carbonate , ZnCO3 . The zinc ion also forms a complex with hydroxide ions . Write a balanced net ionic equation to show why the solubility of ZnCO3 (s) increases in the presence of hydroxide ions and calculate the equilibrium constant for this reaction. For Zn(OH)42- , Kf = 2.9×1015 . Use the pull-down boxes to specify states such as (aq) or (s).
Answer:
The net ionic equation is [tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]
The equilibrium constant is [tex]K = 4.06 *10^{4}[/tex]
Explanation:
From the question we are that
The [tex]K_f = 2.9 *10^{15 }[/tex]
The ionic equation is chemical represented as
Step 1
[tex]ZnCO_3 _{(s)}[/tex] ⇔ [tex]Zn^{2+} _{aq} + CO_3^{2-} _{aq}[/tex] The solubility product constant for stage is [tex]K_{sp} = 1.4*10^{-11}[/tex]
Step 2
[tex]Zn^{2+} _{(aq)} + 4 0H^{-} _{(aq)}[/tex] ⇔ [tex][Zn(OH_4)]^{2-} _{(aq)}[/tex] The formation constant for this step is given as [tex]K_f = 2.9 *10^{15 }[/tex]
The net reaction is
[tex]ZnCO_3 _{(s)} + 4 OH^{-}_{(aq)} \to [Zn(OH)_4]^{2-} _{(aq)} + CO_3^{2-} _{(aq)}[/tex]
The equilibrium constant is mathematically evaluated as
[tex]K = K_{sp} * K_f[/tex]
substituting values
[tex]K = 1.4*10^{-11} * 2.9 *10^{15}[/tex]
[tex]K = 4.06 *10^{4}[/tex]
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
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!
An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as substrate, it was completely oxidized to CO2. However, when undecanoic acid (11:0) was added as substrate, incomplete oxidation occurred unless he bubbled CO2 through the reaction mixture. The addition of the protein avidin, which binds tightly to biotin, prevented the complete oxidation of undecanoic acid even in the presence of CO2, although it had no effect on palmitate oxidation. Explain these observations in light of what you know of fatty acid oxidation reactions.
Answer:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.
Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin. Palmitate oxidation however, does not involve carboxylation.
Explanation:
Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.
Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.
The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme. The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.
Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.
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
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.
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
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.
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
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.
Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set-up
The question is incomplete; the complete question is;
Before running a titration, you calculate the expected endpoint. However, when performing the experiment, you pass the expected endpoint with no visible color change. What is the most likely problem with the titration set- up? Select one
a) There was an air bubble in the burette tip.
b) There is not enough indicator in the analyte.
c) The burette tip is leaking titrant into the analyte.
d) The analyte solution is being stirred too quickly
Answer:
a) There was an air bubble in the burette tip.
Explanation:
Titration involves the determination of the concentration of a solution by measuring the volumes of reactants used in the reaction. The concentration of one of the species must be known while the concentration of the other specie is to be determined by the volumetric analysis.
However, if there are air bubbles at the tip of the burette, this will cause less volume of titrant to be delivered from the burette than expected. Hence, the analyst may think that a certain volume of titrant has been delivered while in reality, a lesser volume was actually delivered due to the air bubbles present. Hence, the analyst may pass the expected endpoint without any colour change because of this problem.
From the available options to the question:
a) There was an air bubble in the burette tip.
b) There is not enough indicator in the analyte.
c) The burette tip is leaking titrant into the analyte.
d) The analyte solution is being stirred too quickly
The most likely problem with the titration setup that could make one to pass the expected endpoint with no visible color change would be if there is not enough indicator in the analyte. The correct option would be B.
A suitable quantity (in drops) of the indicator should be added to the analyte in the conical flask before carrying out a titration. The color of indicators changes quickly near their pKa.
If too few drops of the indicator is used, the color change will be too faint to be obvious and the endpoint will be exceeded. If too many drops of the indicator is used, the final pH of the reaction would be affected and the titer value will be inaccurate.
In this case, the expected endpoint has been exceeded without any color change. The most likely problem would, therefore, be that there is not enough indicator in the analyte.
More on indicators can be found here: https://brainly.com/question/4050911
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.
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]
A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.9 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g·K) and that the densities of the reactant solutions are both 1.07 g/mL. Enter your answer to three significant figures in units of kJ/mol.
Answer:
The correct answer to the following question will be "90.6 kJ/mol".
Explanation:
The total reactant solution will be:
[tex](31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g[/tex]
The produced energy will be:
[tex]=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K[/tex]
[tex]=450.35\times 3.2[/tex]
[tex]=1441.12 \ J[/tex]
The reaction will be:
⇒ [tex]HCl+NaOH \rightarrow NaCl+H_{2}O[/tex]
Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.
Therefore, the moles generated by NaCl will indeed be:
= [tex](\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}[/tex]
= [tex]0.0318\times 0.500[/tex]
= [tex]0.0159 \ mole \ of \ NaCl[/tex]
Now,
= [tex]\frac{1441.12 \ J}{0.0159 \ moles \ NaCl}[/tex]
= [tex]906364.7[/tex]
= [tex]90.6 \ KJ/mol \ NaCl[/tex]
Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of:
Here is the complete question.
Be sure to answer all parts. Three 8−L flasks, fixed with pressure gauges and small valves, each contain 4 g of gas at 276 K. Flask A contains He, flask B contains CH4, and flask C contains H2. Rank the flask contents in terms of: the following properties. (Use the notation >, <, or =, for example B=C>A.)
(a) pressure
(b) average molecular kinetic energy
(c) diffusion rate after the valve is opened
(d) total kinetic energy of the molecules
Answer:
Explanation:
Given that:
Three flask A,B, C:
contains a volume of 8-L
mass m = 4g &;
Temperature = 276 K
Flask A = He
Flask B = H₂
Flask C = CH₄
a) From the ideal gas equation:
PV = nRT
where;
n = number of moles = mass (m)/molar mass (mm)
Then:
PV = m/mm RT
If T ,m and V are constant for the three flasks ; then
P ∝ 1/mm
As such ; the smaller the molar mass the larger the pressure.
Now; since the molecular weight of CH₄ is greater than He and H₂ and also between He and H₂, He has an higher molecular weight .
Then the order of pressure in the flask is :
[tex]\mathbf{P_B >P_A>P_C}[/tex]
where :
[tex]P_A[/tex] = pressure in the flask A
[tex]P_B[/tex] = pressure in the flask B
[tex]P_C[/tex]= Pressure in the flask C
b)
average molecular kinetic energy
We all know that the average molecular kinetic energy varies directly proportional to the temperature.
Thus; the given temperature = 276 K
∴ The order of the average molecular kinetic energy is [tex]\mathbf{K.E_A =K.E_B =K.E_C}[/tex]
c)
The rate of diffusion of gas is inversely proportional to the square root of it density . Here the density is given in relation to their molar mass.
So;
rate of diffusion ∝ [tex]\dfrac{1}{\sqrt{mm} }[/tex]
where;
[tex]D_A[/tex] = rate of diffusion in flask A
[tex]D_B[/tex] = rate of diffusion in flask B
[tex]D_C[/tex] = rate of diffusion in flask C
Thus; the order of the rate of diffusion = [tex]D_B[/tex] > [tex]D_A[/tex] > [tex]D_C[/tex]
d) total kinetic energy of the molecules .
The kinetic energy deals with how the speed of particles of a substance determines how fast the substances will diffuse in a given set of condition.
The the order of the total kinetic energy depends on the molecular speed
Thus; the order of the total kinetic energy for the three flask is as follows:
[tex]\mathbf{ K.E_B>K.E_A>K.E_C}[/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.
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.
describe how would you use chromatography to show whether blue ink contains a single purple dye or a mixture of dyes
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
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