Answer: 3390
Explanation:
Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for [tex]K_{p}[/tex].
[tex]K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}] }[/tex]
[tex]K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390[/tex]
Consider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.
Answer:
Consider each of the following molecules in turn: (a) dimethyl ether, (CH3)2O; (b) trimethylamine, (CH3)3N; (c) trimethylboron, (CH3)3B; and (d) carbon dioxide (CO2). Describe the hybridization state of the central atom (i.e., O, N, B, or C) of each molecule, tell what bond angles you would expect at the central atom, and state whether the molecule would have a dipole moment.
The element silver has an atomic weight of 108 and consists of two stable isotopes silver-107 and silver-109. The isotope silver-107 has a mass of 107 amu and a percent natural abundance of 51.8 %. The isotope silver-109 has a percent natural abundance of 48.2 %. What is the mass of silver-109
Answer:
109
Explanation:
Let silver-107 be isotope A
Let silver-109 be isotope B
Let silver-107 abundance be A%
Let silver-109 abundance be B%
The following data were obtained from the question:
Atomic weight of silver = 108
Mass of isotope A (silver-107) = 107
Abundance of isotope A (silver-107) = 51.8%
Abundance of isotope B (silver-109) = 48.2%
Mass of isotope B (silver-109) =?
Now, we shall determine the mass silver-109 as follow:
Atomic weight = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]
108 = [(107 x 51.8)/100] + [(Mass of B x 48.2)/100]
108 = 55.426 + (Mass of B x 0.482)
Collect like terms
Mass of B x 0.482 = 108 – 55.426
Mass of B x 0.482 = 52.574
Divide both side by 0.482
Mass of B = 52.574/0.482
Mass of B = 109
Therefore, the mass of silver-109 is 109.
AMMONIUM CARBONATE
5. How many grams of nitrogen (N) are in a mass of ammonium carbonate that contains
1.23x10^23 carbon atoms?
Answer:
Zero
Explanation:
Hello,
The question require us to calculate the mass of nitrogen present in aluminium carbonate.
This can easily be calculated using Avogadro's number as a constant with some minor calculations but however in this case, we can't because there's no single atom of nitrogen present in aluminium carbonate hence we can't calculate the mass of nitrogen present in it.
Chemical formula of aluminium carbonate = Al₂(CO₃)₃.
From the above chemical formula, we can see that there's no single atom of nitrogen present in the formula hence the mass of nitrogen present in aluminium carbonate that contains 1.23×10²³ carbon atoms is zero.
Spell out the full name of the compound.
Explanation:
Step one look for the longest chain of carbon atoms
Longest chain is 7 C atoms
Step 2 look for double bonds or others functional groups
it is present in 3rd carbon
Therefore IUPAC name is 3-heptene
From point of stereochemistry it can also be written as trans-3-heptene as the hydrogens are placed in opposite side of the C=C bond.
Hope this helps...
The compound name is: trans 3-heptene
What is molecular compound?Molecular compounds are inorganic compounds that take the form of discrete molecules.
Looking at the given compound:
1. We need to look for the highest carbon chain, So in this compound the highest carbon chain is of 7 carbon atoms.
2. This compound also has a double bond in between that is present at the third carbon which can be detected by numbering the carbon in a order where the lowest number will come over a double bond.
3. Lastly, we can derive the name for this compound as hept-3-ene or 3-heptene.
4. Also, there is one more thing to notice here which is the position of two hydrogen that are present as substituents since they are placed opposite to each other thus we can name it as trans 3-heptene.
Thus, the compound name is: trans 3-heptene.
Find more information about Compound here:
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QUESTION 11
Which of the following compounds has the highest boiling poid?
O CH3CH2CH2COOH
O CH3OCH2CH2CH3
O CH3CH2CH2OH
O CH3CH2CH2CHO
Answer:
CH3CH2CH2COOH
Explanation:
Both carboxylic acids and alcohols posses hydrogen bonding. The difference between the two lies in the strength of the hydrogen bonding and the structure of the molecules.
Alcohols predominantly form linear hydrogen bonds in which the dipole of the -OH group of one molecule interacts with that of another molecule. This gives a linear arrangement of hydrogen bonded intermolecular interactions which significantly impacts the boiling point of alcohols.
However, the carboxylic acids posses the carbonyl (C=O) which is more polar and interacts more effectively with the -OH bond to form dimmer species. These dimmers have a much higher boiling point than the corresponding alcohols due to stronger hydrogen bonds. Hence CH3CH2CH2COOH has a greater boiling point than CH3CH2CH2OH.
The other compounds in the options do not posses hydrogen bonds hence they will have much lower boiling points.
Which metal will spontaneously react with Zn2+
(aq),
but will not spontaneously react with Mg2+
(aq)
according to the activity series
The options of the question is not given so the options are;
(1) Mn(s)
(2) Cu(s)
(3) Ni(s)
(4) Ba(s)
Answer: Mn(s)
Explanation:
The spontaneous reaction depends on the Eo value. The positive Eo value are spontaneous and the negative Eo values are not spontaneous.
so, here the Eo values are:
Eo Zn2+/Zn = -0.763v
Eo Mg2+/Mg = 2.37v
Eo Mn2+/Mn = -1.18v
Therefore, Eo cell (with Zn as one of the half-cell) = Eo Zn2+/Zn - Eo Mn2+/Mn
= -0.763 - (-1.18)
= 0.417v
Whereas, Eo cell (with Mg as one of the half-cell) = Eo Mg2+/Zn - Eo Mn2+/Mn
= -2.37 - (-1.18) = -1.19v
Thus, Mn(s) metal will spontaneously react because it has a positive Eo value and Zn2+(aq), but will not spontaneously react with Mg2+(aq) because it has negative Eo value.
Pre-Lab Study Questions / 9
1. Where are the valence electrons in an atom?
valen
2. How are positive and negative ions fomed?
3. Why are electrons shared in molecular compounds?
Answer:
See detailed answer with explanation below.
Explanation:
Valence electrons are electrons found on the outermost shell of an atom. They are the electrons in an atom that participate in chemical combination. Recall that the outermost shell of an atom is also referred to as its valence shell. Let us consider an example; if we look at the atom, sodium-11, its electronic configuration is 2,8,1. The last one electron is the valence electron of sodium which is found in its outermost or valence shell.
Positive ions are formed when electrons are lost from the valence shell of an atom. For instance, if the outermost electron in sodium is lost, we now form the sodium ion Na^+ which is a positive ion. Positive ions possess less number of electrons compared to their corresponding atoms.
Negative ions are formed when one or more electrons is added to the valence shell of an atom. A negative ion possesses more electrons than its corresponding atom. For example, chlorine(Cl) contains 17 electrons but the chloride ion (Cl^-) contains 18 electrons.
In molecular compounds, a bond is formed when two electrons are shared between the bonding atoms. Each bonding atom may contribute one of the shared electrons (ordinary covalent bond) or one of the bonding atoms may provide the both shared electrons (coordinate covalent bond). The shared pair may be located at an equidistant position to the nucleus of both atoms. Similarly, the electron may be drawn closer to the nucleus of one atom than the other (polar covalent bond) depending on the electro negativity of the two bonding atoms.
The electrons are shared in order to complete the octet of each atom by so doing, the both bonding atoms now obey the octet rule. For example, two chlorine atoms may come together to form a covalent bond in which each chlorine atom has an octet of electrons on its outermost shell.
What force is needed to accelerate a truck with mass 2,000 kg at a rate of 5.0
m/s22
O A. 1,000 N
B. 20,000 N
C. 2,000 N
O D, 10,000 14
Answer:
D
Explanation:
[tex]F=ma \\\\F=2000\cdot 5=10,000N[/tex]
Hope this helps!
How many elements make up most of the ocean?
Answer:
Hydrogen and oxygen
Explanation:
Answer:
eight
Explanation:
the elements of course are the following
hydrogen
oxygen
calcium
sodium
sulfur
magnesium
chlorine
potassium
i hope this helps you
A solid is dissolved in a liquid, and over time a solid forms again. How can
you confirm the type of change that took place?
A. Testing the new solid to show that its properties are the same as
the starting solid would confirm that a physical change took
place.
B. The solid dissolving in a liquid is confirmation that a chemical
change took place.
C. The solid forming from the liquid is confirmation that a physical
change took place.
D. Showing that the total mass of the solid and liquid changed would
confirm that a chemical change took place.
Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6, 0.5 atm for O2, and 2.0 atm for CO2, and 25 oC: C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced) ΔGfo C2H6 (g) = - 32.89 kJ/mol; ΔGfo CO2 (g) = - 394.4 kJ/mol; ΔGfo H2O (l) = - 237.13 kJ/mol
Answer:
Explanation:
Calculate ΔG (in kJ) for the following reaction at 1.0 atm for C2H6,
0.5 atm for O2, and
2.0 atm for CO2, and
25 oC:
C2H6 (g) + O2 (g) ---> CO2 (g) + H2O (l) (unbalanced)
ΔGfo C2H6 (g) = - 32.89 kJ/mol;
ΔGfo CO2 (g) = - 394.4 kJ/mol;
ΔGfo H2O (l) = - 237.13 kJ/mol
The balance equation of this reaction is
[tex]2C_2H_6 (g) + 7O_2 (g) ---> 4CO_2 (g) + 6H_2O (l)[/tex]
[tex]\Delta G_{rxn}=\sum G^o_f(product)-\sum G^o_f(reactant)[/tex]
[tex]=4G^o_f(CO_2)+6G^o_f(H_2O)-7G^o_f(O_2)-2G^o_f(C_2H_6)\\\\[/tex]
[tex][4(-394.4)+6(-237.13)-7(0)-2(-32.89)]kJ/mol\\\\=-1577.6-1422.78+65.78\\\\=-3000.38+65.78\\\\=-2934.6kJ/mol[/tex]
When an automobile engine starts, the metal parts immediately begin to absorb heat released during the combustion of gasoline. How much heat will be absorbed by a 165 kg iron engine block as the temperature rises from 15.7°C to 95.7°C? (The specific heat of iron is 0.489 J/g·°C.)
Answer:
H = 4,034,250 J
Explanation:
Mass, m = 165kg = 165,000g (Converting to grams)
Initial temperature = 15.7°C
Final temperature = 95.7°C
Temperature change, ΔT = 95.7 - 15.7 = 50°C
Specific heat capacity, c = 0.489 J/g·°C
Heat = ?
All the parameters are related with the equation below;
H = m * c * ΔT
H = 165000 * 0.489 * 50
H = 4,034,250 J
The first solution will be an unbuffered sucrose solution. Calculate the amount of 0.1 M sucrose stock solution and the amount of dH2O needed to prepare 100 mL of unbuffered 20 mM sucrose.
Answer:
20mL of the stock solution and dilute to 100mL
Explanation:
20mM = 0.020M is the concentration of the diluted solution. As you want to prepare this solution from a 0.1M solution, dilution factor is:
0.1M / 0.020M = 5
That means you need to dilute the stock solution 5 times to obtain the 20mM sucrose solution.
As you want to prepare 100mL, you need to add:
100mL / 5 =
20mL of the stock solution and dilute to 100mLGiven the following Fischer projection: Fischer projection for an entantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right. draw the perspective formula of the molecule. Be sure to correctly indicate stereochemistry in your answer.
Answer:
Explanation:
Stereoisomers are two or more atoms that have the same bonding order of atoms but there is a difference spatial arrangement of the atoms in space.
A plane of symmetry divides a molecule into two equal halves.
A chiral stereoisomer are not superimposed on a mirror image , Hence they do not posses a plane of symmetry.
As a result to that. these non-superimposable mirror images are said to be Enantiomers.
However, a Fischer Projection emanates from a two - dimensional figure which is used for presenting a three - dimensional organic molecules.
From the given question;
Fischer projection for an enantiomer of 2-bromo-2,3-dihydroxypropanal with the bromine oriented horizontally to the left and the hydroxide group oriented horizontally to the right.
we can sketch the way the enantiomer of 2-bromo-2,3-dihydroxypropanal can be seen like the one shown below:
CH₂OH
|
|
|
Br -------------|----------------OH
|
|
|
CHO
The objective of this question is to drawn the perspective formula of the molecule.
So , from the attached file below; we can see the perspective formula of the molecule in a well structured 3-D format.
Among three bases, X−, Y−, and Z−, the strongest one is Y−, and the weakest one is Z−. Rank their conjugate acids, HX, HY, and HZ, in order of decreasing strength. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them.
Answer: HZ > HX > HY in order of decreasing strengths.
Explanation: Generally, the rule is that the stronger the acid, the weaker its conjugate base and vice versa; same rule applies for bases and their conjugate acids.
So the weakest base Z- would have the strongest conjugate acid. Consequently, the strongest base Y- would have the weakest conjugate acid.
I hope this was MORE helpful as this is the correct answer.
The ranking of the conjugate acids in order of decreasing strength (i.e from strongest to weakest) is; HZ < HX < HY
First we must know that the stronger a base is, the weaker is it's conjugate acid and the weaker a base is, the stronger is it's conjugate acid.
Therefore, the order of decreasing strength of the conjugate acid is; HZ < HX < HY
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an auto of an element has 17 protons in its nucleus.a) write the electronic configuration of the atom.b)to what period and group does the element belong
Answer:
i hope it will help you
Explanation:
electronic configuration 1s²,2s,²2p^6,3s²3p^6,4s^1
as it has one electron in its valence shell so it is the member of group 1A(ALKALI METALS) and the number of shells is 4 so it is in period 4
The reason for the dramatic decline in the number of measles cases from the 1960s to 2010 in the United States was because the vaccine
Answer:
It was because the vaccine generated actively acquired immunity, that is, inoculation of a portion of the measles virus so that the body forms the antibodies for a second contact and thus can destroy it without triggering the pathology.
Explanation:
Vaccines are methods of active acquired immunity since the antibody is not passively inoculated, it is manufactured by the body with a physiological process once part of the virus is inoculated.
The measles virus most of all affected the lives of infants or newborn children with severe rashes and high fevers that led to death.
Use the link Standard Reduction Potentials. Write net equations for the spontaneous redox reactions that occur during the following or none (all lower case) if there is no extensive reaction. Use the lowest possible coefficients. A) Iron metal is dipped into a Ni2+ solution.B) Copper metal is added to hydrochloric acid.C) A silver wire is added to nitric acid (HNO3).D) Magnesium metal is added to waterE) Chromium metal is added to a solution of Mg2+
Answer:
a) Fe(s) + Ni^2+(aq) ----> Fe^2+(aq) + Ni(s)
b) no reaction
c) no reaction
d) 2Mg(s) + 2H2O(l)-----> 2Mg^2+(aq) + O2(g) +4H^+(aq)
e) no reaction
Explanation:
It is important to say here that the ability of a particular chemical specie to displace another chemical specie is dependent on the relative standard reduction potentials of the species involved.
All the reactions stated above are redox reactions. Let us take reaction E as an example. Mg^2+ has a reduction potential of -2.37 V while Cr^3+ has a reduction potential of -0.74V. Since the reduction potential of magnesium is more negative than that of chromium, there is no reaction when a piece of chromium metal is dipped into a solution of Mg^2+.
Similarly, though metals displace hydrogen gas from dilute acids, metals that are less than hydrogen in the reactivity series cannot do that. This explains why there is no reaction when copper and silver are dipped into dilute acid solutions.
Reaction occurs when iron is dipped into a nickel solution because the reduction potential of Fe^2+ is far more negative than that of Ni^2+.
In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OATP+H2O⟶ADP+PiΔG=+13.8 kJ/molΔG=−30.5 kJ/mol Answer the four questions about the first step of glycolysis. Is reaction 1 spontaneous or nonspontaneous?
Answer: Reaction 1 is non spontaneous.
Explanation:
According to Gibb's equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change
[tex]\Delta S[/tex] = entropy change
T = temperature in Kelvin
When [tex]\Delta G[/tex] = +ve, reaction is non spontaneous
[tex]\Delta G[/tex]= -ve, reaction is spontaneous
[tex]\Delta G[/tex]= 0, reaction is in equilibrium
For the given reaction 1: [tex]glucose+Pi\rightarrow glucose-6-phosphate+H_2O[/tex] [tex]\Delta G=+13.8kJ/mol[/tex]
As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.
Because cylinder with a volume of 3.00dm^3 contains 8.00 moles of oxygen gas at a temperature of 50.0 K. What is the pressure inside the cylinder? ( R=8.31)
Answer:
Option A. 1110 KPa.
Explanation:
The following data were obtained from the question:
Volume (V) = 3 dm³
Number of mole (n) = 8 moles
Temperature (T) = 50K
Gas constant (R) = 8.31 KPa.dm³/Kmol
Pressure (P) =..?
Pressure inside the cylinder can be obtained by using the ideal gas equation as follow:
PV =nRT
P x 3 = 8 x 8.31 x 50
Divide both side by 3
P= (8 x 8.31 x 50) /3
P = 1108 ≈ 1110 KPa
Therefore, the pressure inside the cylinder is 1110 KPa
Nitrogen monoxide is produced by combustion in an automobile engine. For the following reaction, 4.36 grams of nitrogen monoxide are mixed with excess oxygen gas . The reaction yields 5.46 grams of nitrogen dioxide . nitrogen monoxide ( g ) oxygen ( g ) nitrogen dioxide ( g ) What is the theoretical yield of nitrogen dioxide
Answer:
[tex]6.68~g~NO_2[/tex]
Explanation:
We have to start with the combustion reaction:
[tex]NO~+~O_2~->~NO_2[/tex]
Then we can balance the reaction:
[tex]2NO~+~O_2~->~2NO_2[/tex]
If we want to find the theoretical yield, we have to calculate the amount of [tex]NO_2[/tex]. To do this, we have to first convert the 4.36 g of [tex]NO[/tex] to moles [tex]NO[/tex] (using the molar mass 30 g/mol), then we have to convert from moles of [tex]NO[/tex] to moles of [tex]NO_2[/tex] (using the molar ratio) finally, we have to convert from moles of [tex]NO_2[/tex] to grams of [tex]NO_2[/tex] (using the molas mass 46 g/mol), so:
[tex]4.36~g~NO\frac{1~mol~NO}{4.36~g~NO}\frac{2~mol~NO_2}{2~mol~NO}\frac{46~g~NO_2}{1~mol~NO_2}=6.68~g~NO_2[/tex]
I hope it helps!
A certain lightbulb containing argon at 1.20 atm and 18°C is
heated to 85°C at constant volume. Calculate its final pressure
(in atm).
Answer:
certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P 1 T 1 P 2 T 2 ... Ideal Gas Equation 5.4 Charles' law: V T (at constant n and P ) ... Consider a case in which two gases, A and B , are in a container of volume V.
Explanation:
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g) 2.10 g were obtained. What is the percent yield of the reaction?
Answer:
[tex]Y=58.15\%[/tex]
Explanation:
Hello,
For the given chemical reaction:
[tex]Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)[/tex]
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
[tex]n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3[/tex]
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
[tex]n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3[/tex]
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
[tex]m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3[/tex]
Finally, we compute the percent yield with the obtained 2.10 g:
[tex]Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%[/tex]
Best regards.
What is the law of conservation and what happens when two substances at different temperatures come into contact.
Answer:
- Both energy and matter cannot be neither created nor destroyed.
- An equilibrium temperature will be reached.
Explanation:
Hello,
In this case, the law of conservation is applied to both matter and energy, and it states that both energy and matter cannot be neither created nor destroyed. Specifically, in chemical reactions, it states that in closed systems, the mass of the reactants equals the mass of the products even when the number of moles change. Moreover, for energy, if two substances at different temperatures come into contact, the hot one will cool down and the cold one will heat up until an equilibrium temperature so the energy lost by the hot one is gained by the cold one, which accounts for the transformation of energy.
Best regards.
For each of the following pairs of elements
(1C and N2) (1Ar and Br2)
pick the atom with
a. more favorable (exothermic) electron affinity.
b. higher ionization energy.
c. larger size.
How do you even go about do this?
Explanation:
Electron affinity is defined as the energy released by the addition of an electron to any gaseous atom. Electron affinity of an atom depends on the electronic configuration.
a).The carbon has vacant p-orbital and nitrogen has half-filled configuration which is more stable. Therefore, one electron can be easily added to carbon whereas nitrogen having more stable configuration releases more amount of energy on adding one electron. Therefore, nitrogen has more electron affinity than carbon.
The bromine has vacant p-orbital whereas argon has filled orbital which is most stable. Therefore, one electron can be easily added to bromine whereas argon having more stable configuration releases more amount of energy on adding one electron.Therefore, argon has more electron affinity than bromine.
Answer:
1. a. C; b. N; c. C; 2. a. Br; b. Ar; c. Br
Explanation:
Use your Periodic Table and follow the trends in atomic properties (Fig. 1).
Electron affinity increases from left to right and from bottom to top.
The elements with the most exothermic EA are at the upper right corner
Exceptions are the noble gases (group 18) and the pnictogens (group 18).
The elements of Group 18 have a complete octet and have no tendency to accept electrons.
The elements of Group 15 have half-filled p subshells. They are more stable than the elements immediately preceding them, so they are less exothermic when adding an electron.
Ionization energy increases from left to right and from bottom to top.
The atoms with the highest IE are at the upper right corner.
Atomic size increases from right to left and from bottom to top.
The largest atoms are on the lower-left corner.
1. C vs N
(a) EA: C. N is a Group 15 element
(b) IE: N. N is further to the right.
(c) Size: C. C is further to the left.
2. Ar vs Br
(a) EA: Br. Ar is a noble gas.
(b) IE: Ar. Ar is further to the right.
(c) Size: Br. Br is nearer to the bottom.
The lock-and-key model and the induced-fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Following are several statements concerning enzyme and substrate interaction. Indicate whether each statement is part of the lock-and-key model, the induced-fit model, or is common to both models.
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions
Answer:
The lock-and-key model:
c. Enzyme active site has a rigid structure complementary
The induced-fit model:
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate.
Common to both The lock-and-key model and The induced-fit model:
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex.
d. Substrate binds to the enzyme through non-covalent interactions
Explanation:
Generally, the catalytic power of enzymes are due to transient covalent bonds formed between an enzyme's catalytic functional group and a substrate as well as non-covalent interactions between substrate and enzyme which lowers the activation energy of the reaction. This applies to both the lock-and-key model as well as induced-fit mode of enzyme catalysis.
The lock and key model of enzyme catalysis and specificity proposes that enzymes are structurally complementary to their substrates such that they fit like a lock and key. This complementary nature of the enzyme and its substrates ensures that only a substrate that is complementary to the enzyme's active site can bind to it for catalysis to proceed. this is known as the specificity of an enzyme to a particular substrate.
The induced-fit mode proposes that binding of substrate to the active site of an enzyme induces conformational changes in the enzyme which better positions various functional groups on the enzyme into the proper position to catalyse the reaction.
Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess? 8. How many grams of F2 are needed to react with 3.50 grams of Cl2?
Answer:
5) 1.147 moles Cl2
6) 12.57 grams ClF3
7) 339.10 grams ClF3
8) 5.63 grams F2
Explanation:
Step 1: Data given
Number of moles F2 = 3.44 moles
Molar mass F2 = 38.00 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles F2
For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3
For 3.44 moles F2 we'll need 3.44/3 = 1.147 moles Cl2
Step 1: Data given
Number of moles F2 = 0.204 moles
Molar mass F2 = 38.00 g/mol
Molar mass ClF3 = 92.448 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles ClF3
For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3
For 0.204 moles F2 we'll have 2/3 * 0.204 = 0.136 moles
Step 4: Calculate mass ClF3
Mass ClF3 = Moles ClF3 * molar mass ClF3
Mass ClF3 = 0.136 moles * 92.448 g/mol
Mass ClF3 = 12.57 grams ClF3
Step 1: Data given
Mass of Cl2 = 130.0 grams
Molar mass F2 = 38.00 g/mol
Molar mass ClF3 = 92.448 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles Cl2
Moles Cl2 = mass Cl2 / molar mass Cl2
Moles Cl2 = 130.0 grams / 70.9 g/mol
Moles Cl2 = 1.834 moles
Step 4: Calculate moles
For 1 mol Cl2 we need 3 moles F2 to produce 2 moles ClF3
For 1.834 moles Cl2 e'll have 2*1.834 = 3.668 moles ClF3
Step 5: Calculate mass ClF3
Mass ClF3 = Moles ClF3 * molar mass ClF3
Mass ClF3 = 3.668 moles * 92.448 g/mol
Mass ClF3 = 339.10 grams ClF3
Step 1: Data given
Mass of Cl2 = 3.50 grams
Molar mass F2 = 38.00 g/mol
Molar mass ClF3 = 92.448 g/mol
Step 2: The balanced equation
Cl2 + 3F2 → 2ClF3
Step 3: Calculate moles Cl2
Moles Cl2 = Mass Cl2 / molar mass Cl2
Moles Cl2 = 3.50 grams / 70.9 g/mol
Moles Cl2 = 0.0494 moles
Step 4: Calculate moles F2
For 1 mol Cl2 we need 3 moles F2
For 0.0494 moles we need 3*0.0494 = 0.1482 moles
Step 5: Calculate mass F2
Mass F2 = moles F2 * molar mass F2
Mass F2 = 0.1482 moles * 38.00 g/mol
Mass F2 = 5.63 grams F2
The amount of calcium in a 15.0-g sample was determined by converting the calcium to calcium oxalate, CaC2O4. The CaC2O4 weighed 40.3 g. What is the percent of calcium in the original sample
Answer:
128 gram of CaC2O4 contains 40 gram of Calcium
40.3 gram of CaC2O4 cotnains = 40*40.3/128 = 12.59 gram of Calcium
out of 15 gram 12.59 gram is Calcaim that means around 50% of orginal sample has Calcium
Nitrosyl bromide, NOBr, is formed from NO and Br2.
2 NO(g) + Br2(g) 2 NOBr(g)
Experiments show the reaction is second order in NO and first order in Br2.
How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
a) the rate will double
b) the rate will remain unchanged
c) the rate will triple
Answer:
C ) The rate will triple
1. Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estimating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).
Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M
Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
Drag the appropriate items to their respective bins.
1. [OH−] = 6×10−12 M
2. [OH−] = 9×10−9 M
3. [OH−] = 8×10−10 M
4. [OH−] = 7×10−13 M
5. [OH−] = 2×10−2 M
6. [OH−] = 9×10−4 M
7. [OH−] = 5×10−5 M
8. [OH−] = 1×10−7 M
A. Acidic
B. Neutral
C. Basic
2. A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [OH−] of the solution.
3. A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [H3O+] of the solution.
Answer:
Question 1.
1. [OH−] = 6×10−12 M is less than 1 * 10⁻⁷, therefore is acidic.
2. [OH−] = 9×10−9 M is less than 1 * 10⁻⁷, therefore is acidic.
3. [OH−] = 8×10−10 M is less than 1 * 10⁻⁷, therefore is acidic.
4. [OH−] = 7×10−13 M is less than 1 * 10⁻⁷, therefore is acidic.
5. [OH−] = 2×10−2 M is greater than 1 * 10⁻⁷, therefore is basic.
6. [OH−] = 9×10−4 M is greater than 1 * 10⁻⁷, therefore is basic.
7. [OH−] = 5×10−5 M is greater than 1 * 10⁻⁷, therefore is basic.
8. [OH−] = 1×10−7 M is equal to 1 * 10⁻⁷, therefore is neutral
Question 2:
[OH⁻] = 1.92 * 10⁻⁸ M
Question 3:
[H₃O⁺] = 3.70 * 10⁻¹¹ M
Explanation:
The ion product constant of water Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.
A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.
1. [OH−] = 6×10−12 M is less than 1 * 10⁻⁷, therefore is acidic.
2. [OH−] = 9×10−9 M is less than 1 * 10⁻⁷, therefore is acidic.
3. [OH−] = 8×10−10 M is less than 1 * 10⁻⁷, therefore is acidic.
4. [OH−] = 7×10−13 M is less than 1 * 10⁻⁷, therefore is acidic.
5. [OH−] = 2×10−2 M is greater than 1 * 10⁻⁷, therefore is basic.
6. [OH−] = 9×10−4 M is greater than 1 * 10⁻⁷, therefore is basic.
7. [OH−] = 5×10−5 M is greater than 1 * 10⁻⁷, therefore is basic.
8. [OH−] = 1×10−7 M is equal to 1 * 10⁻⁷, therefore is neutral
Question 2:
Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]
[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M
[OH⁻] = 1.92 * 10⁻⁸ M
Question 3:
Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²
[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴
[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]
[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M
[H₃O⁺] = 3.70 * 10⁻¹¹ M