Answer:
Turnover number = 113182s⁻¹
Explanation:
Turnover number is a concept used to know the activity of an enzyme. The higher turnover number, the higher activity of the enzime.
Turnover number is defined as:
Turnover number = Rmax / [E]t
As Rmax is 249 μmol⋅L⁻¹⋅s⁻¹ and [E]t is 2.20 nmol⋅L⁻¹
As 1μmol = 1000nmol, Rmax = 249000 nmol⋅L⁻¹⋅s⁻¹. Replacing:
Turnover number = 249000 nmol⋅L⁻¹⋅s⁻¹ / 2.20 nmol⋅L⁻¹
Turnover number = 113182s⁻¹Consider 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]
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:
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
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 .
A new non-electrolyte molecular is discovered. When 241 mg of the molecular is dissolved in 250.0 mL of water, it has an osmotic pressure of 0.072 atm at 25 oC.What is the molar mass of the molecular
Answer:
Molar mass = 32.62g/mol
Explanation:
Mass = 24mg
Volume = 250.0ml
Osmotic pressure = 0.072 atm
Temperature = 25 oC = 298 K (Converting to Kelvin temperature)
Mass concentration = Mass / Volume
Mass concentration = 24 mg / 250 ml = 0.096 mg/ml = 0.096 g / L
This means that in 1L, there are 0.096g of the non electrolyte.
Molar mass = Mass / Moles
Finding the Molarity using osmotic pressure and temperature
p = MRT
Where M = n / v
M = p / RT
M = 0.072 / (0.0821 L atm/mol K) (298K)
M = 0.072 / 24.4658
M = 0.002943M
Find the moles of solute from molarity by multiplying by the liters of solution. Since our calculation is base on 1L,
moles = Molarity * Volume
moles = 0.002943M * 1
moles = 0.002943 mol
Molar mass = Mass / Moles
Molar mass = 0.096g / 0.002943 mol
Molar mass = 32.62g/mol
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)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]
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.
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:
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.
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.
You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.2750 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator).
Required:
a. What is the molarity of the acetic acid solution?
b. What is the percentage of acetic acid in the solution?
Answer:
a. 0.393M CH₃COOH.
b. 2.360% of acetic acid in the solution
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺
That means 1 mole of acid reacts per mole of NaOH.
Moles of NaOH to reach the equivalence point are:
35.75mL = 0.03575L × (0.2750mol / L) = 9.831x10⁻³ moles of NaOH
As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.
a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:
9.831x10⁻³ moles / 0.02500L =
0.393M CH₃COOHb. Molar mass of acetic acid is 60g/mol. The mass of 9.831x10⁻³ moles is:
9.831x10⁻³ moles ₓ (60g / mol) = 0.590g of CH₃COOH.
As volume of the solution is 25.00mL, the percentage of acetic acid is:
(0.590g CH₃COOH / 25.00mL) ₓ 100 =
2.360% of acetic acid in the solutionAn 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]
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.
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 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]
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.
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.
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.
The 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
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]
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!
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.
The carbon-carbon bond length in ethylene is ________ than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is ________ the HCH bond angle in ethane
Answer:
shorter
longer
Explanation:
The carbon-carbon bond length in ethylene is shorter than the carbon-carbon bond length in ethane, and the HCH bond angle in ethylene is longer the HCH bond angle in ethane.
The objective of this question is to let us understand the concept of Bond Length and Bond angle among the unsaturated aliphatic hydrocarbons (i.e alkanes, alkenes and alkynes).
The variation in bond angles of unsaturated aliphatic hydrocarbons can be explained by two concepts; The valence shell electron pair repulsion (VSEPR) model and hybridization.
The VSEPR model determines the total number of electron pairs surrounding the central atom of a species. The total number of electron pairs consist of the bond pairs and lone pairs. All the electron pairs( lie charge ) will then orient themselves in such a way to minimize the electrostatic repulsion between them.
As the number of the lone pairs increases from zero to 2 ; the bond angles diminish progressively.
However;
Hybridization is the mixing or blending of two or more pure atomic orbitals (s,p and d) to form two or more hybrid atomic orbitals that are identical in shape and energy . e.g sp, sp² , sp³ hybrid orbitals etc .
The shape of the geometry of this compound hence determines their bond angle.
The shape of the geometry of ethane is tetrahedral which is 109.5° in bond angle while that of ethylene is trigonal planar which is 120°.
This is why the HCH bond angle in ethylene is longer the HCH bond angle in ethane .
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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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:
what volume at stp of (a)o2 (b)H2 (c)Cl2 gas will be liberated during electrolysis by the passage of 48250c of electricity,molar volume=22.4
The volume of o₂ 2.8 liters or 2.8 dm³.
The volume of H₂ 2.8 liters or 5.6dm³.
The volume of H₂ 2.8 liters or 5.6 dm³.
What s electrolysis?
Electrolysis is the process by which electric current is passed through an electrolyte to effect a chemical change.
Through this chemical change in which the substance loses gets oxidized or gains an electron reduced.
calculations:-for a divalent molecule O2 4*96500, C = 1 mole of gas evolved.
∴48250c =48250/(4*96500)=0.125mole
volvme = 0.125*22.4= 2.8 litre.
for monovalent cl and H2 molecules, 2*96500c = 1 mole of gas is evolved.
∴48250c =48250/(2*96500)=0.25mole
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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]
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.
Apart from carbon, name other element which exhibits (slow) allotropy
Answer:
Apart from carbon, Sulfur also shows allotropy and has the following allotropes.
1) Mono clinic Sulfur
2) Rhombic Sulfur