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!
475 grams of solid calcium oxide reacts with water vapor to form solid calcium hydroxide.Calculate the heat of reaction.
(A) - 221.1 kcal
(B) + 290.8 kcal
C) - 290.8 kcal
(D) + 324.6 kcal
(E) + 221.1 kcal
Answer:
(A) - 221.1 kcal
Explanation:
Based in the reaction:
CaO(s) + H₂O(g) → Ca(OH)₂ ΔH = -109kJ/mol
When 1 mole of CaO reacts per mole of water vapor producing calcium hydroxide there are released -109kJ
475g of CaO (Molar mass CaO: 56.08g/mol) are:
475g CaO × (1mol / 56.08g) = 8.47 moles of CaO
As 1 mole of CaO in reaction release -109kJ, 8.47 moles release:
8.47 mol CaO × (-109 kJ / 1 mol CaO) = -923.2kJ are released
As 1 kCal = 4.184kJ:
-923.2kJ × (1kCal / 4.184kJ) =
-220.7kCal ≈ (A) - 221.1 kcalThe 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
g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass. Must show your work on scratch paper to receive credit.
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
s 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.
Which is regarding enzyme inhbition
Explanation:
An enzyme inhibitor is a molecule that binds to an enzyme and decreases its activity. ... Since blocking an enzyme's activity can kill a pathogen or correct a metabolic imbalance, many drugs are enzyme inhibitors. They are also used in pesticides.
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
What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?Cu + 2AgNO3→Cu(NO3)2+ 2Ag
Answer: 323.61 g of [tex]Ag[/tex] will be produced
Explanation:
The given balanced chemical reaction is :
[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]
According to stoichiometry :
2 moles of [tex]AgNO_3[/tex] require 1 mole of [tex]Cu[/tex]
Thus 3.00 moles of [tex]AgNO_3[/tex] will require=[tex]\frac{1}{2}\times 3.00=1.50moles[/tex] of [tex]Cu[/tex]
Thus [tex]AgNO_3[/tex] is the limiting reagent as it limits the formation of product.
As 2 moles of [tex]AgNO_3[/tex] give = 2 moles of [tex]Ag[/tex]
Thus 3.00 moles of [tex]AgNO_3[/tex] give =[tex]\frac{2}{2}\times 3.00=3.00moles[/tex] of [tex]Ag[/tex]
Mass of [tex]Ag=moles\times {\text {Molar mass}}=3.00moles\times 107.87g/mol=323.61g[/tex]
Thus 323.61 g of [tex]Ag[/tex] will be produced from the given moles of both reactants.
Answer:
B). 2KNO3 Is your answer
Explanation:
Which of the following aqueous solutions are good buffer systems?
0.31 M ammonium bromide + 0.39 M ammonia
0.31 M nitrous acid + 0.25 M potassium nitrite
0.21 M perchloric acid + 0.21 M potassium perchlorate
0.16 M potassium cyanide + 0.21 M hydrocyanic acid
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite
0.13 M nitrous acid + 0.12 M potassium nitrite
0.15 M potassium hydroxide + 0.22 M potassium bromide
0.23 M hydrobromic acid + 0.20 M potassium bromide
0.34 M calcium iodide + 0.29 M potassium iodide
0.33 M ammonia + 0.30 M sodium hydroxide
0.20 M nitrous acid + 0.18 M potassium nitrite
0.30 M ammonia + 0.34 M ammonium bromide
0.29 M hydrobromic acid + 0.22 M sodium bromide
0.17 M calcium hydroxide + 0.28 M calcium bromide
0.34 M potassium iodide + 0.27 M potassium bromide
Answer:
Answers are in the explanation.
Explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa. Having this in mind:
0.31 M ammonium bromide + 0.39 M ammonia . Is a good buffer system because ammonia is a weak base and its conjugate base, ammonium ion is in the solution.
0.31 M nitrous acid + 0.25 M potassium nitrite . Is a good buffer system because nitrous acid is the weak acid and nitrite ion its conjugate base.
0.21 M perchloric acid + 0.21 M potassium perchlorate . Perchloric acid is a strong acid. Thus, Is not a good buffer system.
0.16 M potassium cyanide + 0.21 M hydrocyanic acid . Hydrocyanic acid is a weak acid and cyanide ion is its conjugate base. Is a good buffer system.
0.14 M hypochlorous acid + 0.21 M sodium hypochlorite . Hypochlorous acid is a weak acid and hypochlorite ion its conjugate base. Is a good buffer system.
0.13 M nitrous acid + 0.12 M potassium nitrite . Is a good buffer system as I explained yet.
0.15 M potassium hydroxide + 0.22 M potassium bromide . Potassium hydroxide is a strong base. Is not a good buffer system.
0.23 M hydrobromic acid + 0.20 M potassium bromide . HBr is a strong acid. Is not a good buffer system.
0.34 M calcium iodide + 0.29 M potassium iodide . CaI and KI are both salts, Is not a good buffer system.
0.33 M ammonia + 0.30 M sodium hydroxide . Ammonia is a weak base but its conjugate base ammonium ion is not in solution. Is not a good buffer system.
0.20 M nitrous acid + 0.18 M potassium nitrite . Is a good buffer system.
0.30 M ammonia + 0.34 M ammonium bromide . Ammonia and ammonium in solution, Good buffer system.
0.29 M hydrobromic acid + 0.22 M sodium bromide . HBr is a strong acid, is not a good buffer system.
0.17 M calcium hydroxide + 0.28 M calcium bromide . CaOH is a strong base, is not a good buffer system.
0.34 M potassium iodide + 0.27 M potassium bromide. KI and KBr are both salts, is not a good buffer system.
In supersonic flights, molecules break apart and react chemically. which safety features protect the plane?
Answer:
Explanation:
The heat Shield are materials (usually made of metals) protect us from heat by absoring lots of heat and gradually releasing heat by surrounding air cirucaltion
Answer:
Heat shield
Explanation: Most heat shields consist of one or more layers of stamped metal that are shaped into a shield that is designed to wrap around the exhaust manifold. The shield acts as a barrier and heat sink, preventing the heat from the manifold from reaching any of the components under the hood and potentially causing damage.
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]
In a chemical equation, which symbol should be used to indicate that a substance is in solution? (s)
Answer:
(aq) meaning aqueous solution
Explanation:
hope it helps .
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.
Which describes an effect that ocean currents have on short-term climate change? Ocean currents increase the strength of prevailing winds, which can cool the air and land. Ocean currents can carry cold water, which can cool the air and land. Ocean currents increase hurricane activity, which can raise the temperature of the air and land. Ocean currents can carry warm water, which causes hurricane activity and raises the temperature of the air and land.
Answer: B
Ocean currents can carry cold water, which can cool the air and land.
Explanation:
eet ees wat eet ees
plz mark brainliest
Answer:
B is right
Explanation:
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]
how to use VBT to draw the orbital overlapping diagram
Answer:
from the valence elecrtons configuration is the centre atom.atomised the number of elecrtons pair determine the hybridization .
Explanation:
you can read this note to know the ans
Mathematical relationship between the rate of a chemical reaction and the
concentration of reactants?
Specific Rate Constant
Collision Theory
Rate Law
Activation energy
Answer:
Specific Rate Constant
Explanation:
Mathematical relationship between the rate of a chemical reaction and the
concentration of reactants is Specific Rate Constant.
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
A scientist measures the standard enthalpy change for the following reaction to be -115.5 kJ: CO(g) + Cl2(g)___COCl2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of COCl2(g) is ________ kJ/mol.
Answer:
-226.0kJ = ΔH°f COCl₂(g)
Explanation:
Using Hess' law, it is possible to obtain the enthalpy of formation of a substance from the enthalpy change of a reaction and the other enthalpies of formation involved in the reaction.
For the reaction:
CO(g) + Cl₂(g) → COCl₂(g)
Hess's law is:
ΔHr = -115.5kJ = ΔH°f COCl₂(g) - (ΔH°f CO(g) + ΔH°f Cl₂(g))
ΔH°f CO(g) is -110.5kJ/mol
ΔH°f Cl₂(g) is 0 kJ/mol
Replacing in Hess's law:
-115.5kJ = ΔH°f COCl₂(g) - (-110.5kJ/mol + 0kJ/mol)
-115.5kJ = ΔH°f COCl₂(g) + 110.5kJ
-226.0kJ = ΔH°f COCl₂(g)which proess is part of the carbon cycle
Answer:
The key processes in the carbon cycle are: carbon dioxide from the atmosphere is converted into plant material in the biosphere by photosynthesis.
Explanation:
organisms in the biosphere obtain energy by respiration and so release carbon dioxide that was originally trapped by photosynthesis. ... The carbon becomes part of the .
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 525 mL of a solution that has a concentration of Na ions of 1.10 M
Answer:
31.652g of Na3PO4
Explanation:
We'll begin by calculating the molarity of Na3PO4 solution. This can be achieved as shown below:
Na3PO4 will dessicate in solution as follow:
Na3PO4(aq) —> 3Na+(aq) + PO4³¯(aq)
From the balanced equation above,
1 mole of Na3PO4 produce 3 moles of sodium ion, Na+.
Therefore, xM Na3PO4 will produce 1.10M sodium ion, Na+ i.e
xM Na3PO4 = (1.10 x 1)/3
xM Na3PO4 = 0.367M
Therefore, the molarity of Na3PO4 is 0.367M.
Next, we shall determine the number of mole of Na3PO4 in the solution. This is illustrated below:
Molarity of Na3PO4 = 0.367M
Volume = 525mL = 525/1000 = 0.525L
Mole of Na3PO4 =..?
Molarity = mole /Volume
0.367 = mole /0.525
Cross multiply
Mole of Na3PO4 = 0.367 x 0.525
Mole of Na3PO4 = 0.193 mole.
Finally, we shall convert 0.193 mole of Na3PO4 to grams. This is illustrated below:
Molar mass of Na3PO4 = (23x3) + 31 + (16x4) = 164g/mol
Mole of Na3PO4 = 0.193 mole
Mass of Na3PO4 =.?
Mass = mole x molar mass
Mass of Na3PO4 = 0.193 x 164
Mass of Na3PO4 = 31.652g
Therefore, 31.652g of Na3PO4 is needed to prepare the solution.
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:
Which diagram represents the bonding pattern of metals?
Answer:
there's no image can't help without it sorry
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]
An electron in a 3s3s orbital penetrates into the region occupied by core electrons more than electrons in a 3p3p orbital. An electron in a orbital penetrates into the region occupied by core electrons more than electrons in a orbital. true false
Answer:
True
Explanation:
The penetrating ability of electrons in the orbitals is in the order s > p > d > f
An electron in a 3s orbital is closer to the nucleus than the one in a 3p orbital and as a result, there will be lesser shielding effect on it. This low shielding effect experienced by the 3s electron gives it a high penetration ability and hence will be able to easily penetrate regions occupied by core electrons. Conversely, the 3p orbital is farther away from the nucleus, electrons revolving around it are highly shielded which limits their ability to penetrate regions of core electrons.
Note that the maximum electrons that the s orbital can accommodate is 2 while p orbital can accommodate a maximum of 8.
In a zero order reaction, it takes 342 seconds for 75% of a hypothetical reactant to decompose. Determine the half-life t_{1/2} in units of seconds. Do not enter units with your numerical answer. Numeric Answer:
Answer:
228 s
Explanation:
In a zero order reaction, the formula for the half life is given as;
t1/2 = [A]o / 2k
To obtain the rate constant k, we have to use;
[A] = [A]o - kt
kt = [A]o - [A]
From the question;
it takes 342 seconds for 75% of a hypothetical reactant to decompose.
We have;
t = 324
[A] = 25
[A]o = 100
Upon solving for k we have;
kt = [A]o - [A]
k = ( [A]o - [A] ) / t
k = (100 - 25 ) / 342
k = 75 / 342 = 0.2193
Solving for t1/2;
t1/2 = [A]o / 2k
t1/2 = 100 / 2(0.2193)
t1/2 = 100 / 0.4386 = 228 s
Calculate the temperature and state the appropriate phase(s) (solid, liquid, vapor) for each substance: This temperature is: Fahrenheit A. SubstanceMelting Point (K)Boiling Point (K)Phase(s) Oxygen, O254.7590.19 Methane, CH493.15109.10 Water, H2O273.15373.15
Answer:
Explanation:
At 54.75K melting point, Oxygen is in gas (vapour) phase
At 373.15K boiling point, water is in liquid phase.
At 109.10K boiling point methane is in gas (vapour) phase.
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.
Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass of 10.88 g is added and the volume of the water now reads 49.4 mL. What is the density of the metal in g/mL? Multiple Choice 10.9 1.8 0.250 1.6 0.541
Answer:
1.8g
Explanation:
Initial volume = 43.5ml
Final volume = 49.4ml
Mass = 10.88g
Density = ?
Volume = Final volume - initial volume
= 49.4 - 43.5
= 5.9ml
Density = Mass/volume
Density = 10.88/5.9
= 1.8g/ml
a) What is the common-ion effect?b) Given an example of a salt that can decrease the ionization of HNO2 in solution. List all of the substances (species) that would be present in the solution. (Consider the HNO2, the salt, and the water.)
Answer:
Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.
NaNO2
Explanation:
In order to understand exactly what common ion effect is, let us consider a simple unambiguous example. Assuming I have a solution of an ionic substance that contains a cation A and an anion B, this ionic substance has chemical formula AB. Secondly, I have another ionic distance with cation C and anion B, its chemical formula is CB. Both CB and AB are soluble in water to a certain degree as shown by their respective KSp.
If I dissolve AB in water and form a solution, subsequently, I add solid CB to this solution, the solubility of CB in this solution is found to be lees than the solubility of CB in pure water because of the ion B^- which is common to both substances in solution. We refer to the phenomenon described above as common ion effect.
Common ion effect refers to the decrease in the solubility of a substance in a solution with which it shares a common ion.
If I try to dissolve NaNO2 in a solution of HNO2, the solubility of NaNO2 in the HNO2 solution will be less than its solubility in pure water due to common ion effect. Also, the extent of ionization of HNO2 in a system that already contains NaNO2 will be decreased compared to its extent ionization in pure water. This system described here will contain HNO2, water and NaNO2
Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.
Question:
When 0.500 g of biphenyl (C₁₂H₁₀) undergoes combustion in a bomb calorimeter, the temperature rises from 26.8 °C to 29.5 °C. Part A Find ΔErxn for the combustion of biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/∘C. Express the energy in kilojoules per mole to three significant figures.
Answer;
-4870kJ/mol(3 significant figure)