Answer:
False
Explanation:
The first reaction is;
NO(g) + 1/2O2(g) ---->NO2(g)
K= [NO2]/[NO] [ O2]^1/2
The second reaction is;
2NO(g) + O2(g) ---->2NO2(g)
K'= [NO2]^2/[NO]^2 [O2]
It now follows that;
K'= K^2
Hence the statement in the question is false
How many moles of Fe contains 3.41 x 1023 Fe atoms?
Answer:
[tex]\boxed {\boxed {\sf 0.566 \ mol \ Fe}}[/tex]
Explanation:
We are asked to convert a number of atoms to moles.
We can convert atoms to moles using Avogadro's Number, which is 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are atoms of iron (Fe). There are 6.022 ×10²³ atoms of iron in 1 mole of iron.
We use dimensional analysis to convert atoms to moles. This involves setting up ratios. Use Avogadro's Number and the underlined information to make a ratio.
[tex]\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]
We are converting 3.41 × 10²³ atoms of iron to moles, so we multiply by this value.
[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac {6.022 \times 10^{23}\ atoms \ Fe}{1 \ mol \ Fe}[/tex]
Flip the ratio. It stays equivalent, but it allows the units of atoms of iron to cancel.
[tex]3.41 \times 10^{23} \ atoms \ Fe *\frac{1 \ mol \ Fe} {6.022 \times 10^{23}\ atoms \ Fe}[/tex]
[tex]3.41 \times 10^{23}*\frac{1 \ mol \ Fe} {6.022 \times 10^{23}}[/tex]
[tex]\frac{3.41 \times 10^{23}} {6.022 \times 10^{23}} \ mol \ Fe[/tex]
[tex]0.5662570575\ mol \ Fe[/tex]
The original measure ment of iron atoms ( 3.41 × 10²³ ) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 2 in the ten-thousandths place ( 0.5662570575) tells us to leave the 6 in the thousandths place.
[tex]0.566 \ mol \ Fe[/tex]
3.41 × 10²³ atoms of iron is equal to approximately 0.566 moles of iron.
Many important analgesic compounds are derived from simple aromatic starting materials.
a. True
b. False
The fact that a beam of particles was deflected in the presence of an electric
or magnetic force led J.J. Thomson to conclude that the particles had a(n)
O A. large mass
B. electric charge
O C. negligible mass
O D. neutral charge
Answer:
electric charge
Explanation:
Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.
When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.
A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.
A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.
For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32
Answer:
by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32
2. XC12 is the chloride of metal X. The formulae of its sulphate is
Answer:
XSO₄
Explanation:
XCl₂ is the chloride of metal X. The sum of the charges of the cation and the anion must be zero because the salt is electrically neutral. The charge of the cation of X is:
1 × X + 2 × Cl = 0
1 × X + 2 × (-1) = 0
X = +2
X has a charge +2 and sulphate (SO₄²⁻) a charge -2. The neutral salt they form is XSO₄.
For the following acids of varying concentrations, which are titrated with 0.50 M NaOH, rank the acids in order of least to most volume of base needed to completely neutralize the acid.
a. 0.2M H2C6H5O7
b. 0.2M H2C2O4
Answer:
0.2M H2C6H5O7 < 0.2M H2C2O4
Explanation:
A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.
More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.
Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.
H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.
H₂C₆H₅O₇ < H₂C₂O₄
The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)
Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.
Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.
Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.
Strong acids ionizes completely to produce hydrogen ions.
Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.
In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.
H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23
H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08
Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇
For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.
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Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
21.4g of HBr is the minimum mass that could be left over.
Explanation:
Based on the reaction:
HBr + NaOH → NaBr + H2O
1 mole of HBr reacts per mole of NaOH
To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:
Moles NaOH -40.0g/mol-
17g * (1mol/40.0g) = 0.425 moles NaOH
Moles HBr -Molar mass: 80.91g/mol-
55.8g * (1mol/80.91g) = 0.690 moles HBr
The difference in moles is:
0.690 moles - 0.425 moles =
0.265 moles of HBr could be left over
The mass is:
0.265 moles * (80.91g/mol) =
21.4g of HBr is the minimum mass that could be left over.A person slips over banana pills. Give reason
Answer:
We slip when we step on a banana peel because the inner side of banana peel being smooth and slippery reduces the friction between the sole of our shoe and the surface of road.
Spell out the full name of the compound.
Answer:
it is propane
C3H8 it is propane
The Full name of the given compound is Propane.
What is Propane ?Propane is a three-carbon alkane with the molecular formula C₃H₈.
It is a gas at standard temperature and pressure, but compressible to a transportable liquid.
In the figure given ;
Black balls represents Carbon atomsWhite balls represents Hydrogen atomsIn the given figure, there is single bond present between Carbon and Hydrogen. Hence, The Full name of the given compound is Propane.
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What is the relationship between temperature and kinetic energy?
Answer:
"[Temperature is a measurement of the average kinetic energy of the molecules in an object or a system. Kinetic energy is the energy that an object has because of its motion. The molecules in a substance have a range of kinetic energies because they don't all move at the same speed.]"
Answer:
Temperature is directly proportional to the average translational kinetic energy of molecules in an ideal gas
Explanation:
Avogradro's number is the number of particles in one gram of carbon- 12 atom true or false?
Answer:
True
Explanation:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
A sample of oxygen occupied 621 mL when the pressure increased to 1095.93mm Hg. At constant temperature, what volume did the gas initially occupy when the pressure was 774.29mm Hg?
a.) 879.0
b.) 438.7
c.) 890.2
d.) 1366
Answer:
for you ta bangla de Mayo for bangla ki you would know it has changed an anyone come by with home home but is in an going forward you as as on Saturday to what our current meeting office next weekend too with we before too
Explanation:
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Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B
were placed in a 5.00-liter container. The mole of B at equilibrium was determined to
be 1.00 mol. Calculate K value.
0.060
5.1
25
17
Ugh
Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
FILL IN THE BLANK:
The rate of a reaction is measured by how fast a (Product Or Reactant)
is used up or how fast a
(Reactant Or Product) is formed?
Answer:
the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed
At 25 oC the solubility of chromium(III) iodate is 2.07 x 10-2 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]
Answer:
5.0 × 10⁻⁶
Explanation:
Step 1: Write the balanced equation for the solution of chromium(III) iodate
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
Step 2: Calculate the solubility product constant (Ksp)
To relate Ksp and the solubility (S), we will make an ICE chart.
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product constant is:
Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶
Which of the following releases hormones into your bloodstream?
A. Endocrine system
B. Sympathetic nervous system
C. Lobal system
a
D. Autonomic nervous system
Answer:
answer is A. Endocrine system
Endocrine glands secrete hormones straight into the bloodstream. Hormones help to control many body functions, such as growth, repair and reproduction.
Answer:
A endocrine system
this is the answer
Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.
a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.
1. Increase
2. decrease
3. No effect
Answer:
a. Decrease
b. Increase
c. Increase
d. No effect
Explanation:
Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.
a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease
b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect
c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase
d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase
A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.
B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.
C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.
D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.
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By using photons of specific wavelengths, chemists can dissociate gaseous HI to produce H atoms with certain speeds. When HI dissociates, the H atoms move away rapidly, whereas the heavier I atoms move more slowly. If a photon of 231 nm is used, what is the excess energy (in J) over that needed for dissociation
Answer:
The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J
Explanation:
Given the data in the question;
wavelength of proton λ = 231 nm = 231 × 10⁻⁹ m
we determine the energy of the proton;
E = hc / λ
where h is plank constant ( 6.626 × 10⁻³⁴ JS )
and c is the speed of light ( 3 × 10⁸ m/s )
we substitute
E = [ ( 6.626 × 10⁻³⁴ JS ) × ( 3 × 10⁸ m/s ) ] / [ 231 × 10⁻⁹ m ]
E = 8.61 × 10⁻¹⁹ J
we know that, bond energy for H-I is 295 kJ/mol
so, H = 295 × 10³ J/mol
Now, energy to dissociate HI will be;
⇒ H / N
where N is the Avogadro's number ( 6.023 × 10²³ mol⁻¹ )
energy to dissociate HI = ( 295 × 10³ J/mol ) / ( 6.023 × 10²³ mol⁻¹ )
= 4.898 × 10⁻¹⁹ J
Therefore, Excess energy over dissociation will be;
⇒ ( 8.61 × 10⁻¹⁹ J ) - ( 4.898 × 10⁻¹⁹ J )
= 3.712 × 10⁻¹⁹ J
The excess energy over that needed for dissociation is 3.712 × 10⁻¹⁹ J
Nucleophilic aromatic substitution involves the formation of a resonance-stabilized carbanion intermediate called a Meisenheimer complex as the nucleophile attacks the ring carbon carrying the eventual leaving group.
a. True
b. False
Answer:
True
Explanation:
Aromatic rings undergo nucleophillic substitution reactions in the presence of a electron withdrawing group which stabilizes the Meisenheimer complex.
When the nucleophile attacks the ring carbon atom carrying the eventual leaving group. A resonance-stabilized carbanion intermediate called a Meisenheimer complex is formed.
Subsequent loss of the leaving group from the intermediate complex yields the product of the reaction.
Arrange the compounds by their reactivity toward electrophilic aromatic substitution.
a. Benzene, ethylbenzene, chlorobenzene, nitrobenzene, anisole.
b. Toluene, p-cresol, benzene, p-xylene.
c. Benzene, benzoic acid, phenol, propylbenzene.
d. p-Methylnitrobenzene, 2-chloro-1-methyl-4-nitrobenzene, 1-methyl-2,4-dinitrobenzene, p-chloromethylbenzene.
Answer:
The order of reactivity towards electrophilic susbtitution is shown below:
a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene
b. p-cresol>p-xylene>toluene>benzene
c.Phenol>propylbenzene>benzene>benzoic acid
d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene
Explanation:
Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.
For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.
The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.
Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.
balance equation of potassium sulphate+ water
Answer:
2KHCO
3
+H
2
SO
4
→K
2
SO
4
+2CO
2
+2H
2
O
The standard entropy change of a reaction has a positive value. This reaction results in: Select the correct answer below: a decrease in entropy. an increase in entropy. no entropy change. neither an entropy increase nor decrease.
Explanation:
The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.
Positive entropy means the system has increased its degree of disorderness.
Which of the following ligands is not capable of exhibiting linkage isomerism?
a. NCO-
b. -OH
c. -CN
d. -SCN
Answer:
a
...
........
...........
The Bohr effect:_____.
a. explains through the Bohr model of the atom why Fe2+ will bind O2 in heme but Fe3+ will not.
b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.
c. applies to both myoglobin and hemoglobin.
d. relates [H+] to [CO2].
Answer:
b. contributes to binding of O2 by hemoglobin in lungs and release of O2 from hemoglobin in tissues.
Explanation:
The Bohr effect is a phenomenon described by Christian Bohr. Is an affinity that binds oxygen and hemoglobin and is inversely related to the concentration of carbon dioxide. As CO2 reacts with water and an increase in CO2 results in a decrease in blood ph.Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
a) potassium tetracyanonickelate(II)
b) sodium diamminedicarbonatoruthenate(III)
c) diamminedichloroplatinum(II)
Answer:
a) K2[Ni(CN)4]
b) Na3[Ru(NH3)2(CO3)2]
c) Pt(NH3)2Cl2
Explanation:
Coordination compounds are named in accordance with IUPAC nomenclature.
According to this nomenclature, negative ligands end with the suffix ''ato'' while neutral ligands have no special ending.
The ions written outside the coordination sphere are counter ions. Given the names of the coordination compounds as written in the question, their formulas are provided above.
Consider a hypothetical metal that has a density of 10.6 g/cm3, an atomic weight of 176.8 g/mol, and an atomic radius of 0.130 nm. Compute the atomic packing factor if the unit cell has tetragonal symmetry, values for the a and c lattice parameters are 0.570 and 0.341, respectively.
Answer:
0.3323
Explanation:
GIven that:
Density of the metal = 10.6 g/cm^3
atomic weight = 176.8 g/mol
atmic radius = 0.130 nm
values of a and c = 0.570 nm and 0.341 nm respectively
For us to determine the atomic packing factor, we need to first determine the volume of all spheres (Vs) and the volume of unit cell (Vc).
However, the number of atoms in the unit cell (n) can be computed as:
[tex]n = \dfrac{\rho * V_c *N_A}{A} \\ \\ n = \dfrac{(10.6) * (5.7)^2 (3.41)*(10^{-24}) *(6.022*10^{23})}{176.8}[/tex]
n = 4.0
Thus, the number of atoms in the unit cell is 4
∴
The atomic paking factor (APF) is calculated by using the formula:
[tex]\dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *R^3 }{a^2 *c} \\ \\ \\ \dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *(1.30*10^{-8})^3 }{(5.70*10^{-8})^2 *(3.41*10^{-8})}[/tex]
= 0.3323
In practice, the second law of thermodynamics means that:
a. Systems move from ordered behavior to more random behavior.
b. Systems move from random behavior to more ordered behavior.
c. Systems move between ordered and random behavior patterns based on temperature.
d. Systems are constantly striving to reach equilibrium.
Answer:
Systems move from ordered behavior to more random behavior.
Explanation:
Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)
According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.
During electrophilic aromatic substitution, a resonance-stabilized cation intermediate is formed. Groups, already present on the benzene ring, that direct ortho/para further stabilize this intermediate by participating in the resonance delocalization of the positive charge. Assume that the following group is present on a benzene ring at position 1 and that you are brominating the ring at positon 4. Draw the structure of the resonance contributor that shows this group actively participating in the charge delocalization.
-----OCH3
Answer:
See explanation and image attached
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.
Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.
Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.
The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.
A substance is made up of slow-moving particles that have very little space between them. Based on this information, what can most likely be concluded about this substance? O It is not a gas because its particles do not move continuously. It is a gas because its particles move continuously in a straight line. 0 It is not a gas because its particles do not have large spaces between them. It is a gas because its particles move in many different directions.
Answer:
o
Explanation:
it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.
Why is bromine more electronegative than iodine?
Answer
Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)
