Answer:
carbons that are in different environments
Explanation:
When molecules are asymmetric every carbon will have its own peak since they are all different and will show up with a different ppm shift. If the molecule has symmetry the carbons that are symmetrical (in the same environment) will have the same ppm shift and will therefore show up as one peak.
An example of a molecule with symmetry is isopropanol which has 3 carbons but only two carbon peaks since the two methyl groups are symmetrical.
An example of a molecule with no symmetry is 3-Nitroaniline where the groups coming off of the benzene ring makes each of the 6 carbons be in different environments and there for all 6 carbons will have different ppm shifts. The result is a carbon NMR that has 6 peaks.
I hope this helps. Let me know if anything is unclear.
Carbon NMR spectroscopy produces a spectrum of only carbon-13 nuclei in a sample. The number of carbon signals in the spectrum corresponds to the number of different number of carbon in the molecule.
What is spectroscopy ?The study of spectroscopy involves measuring and analyzing the electromagnetic spectra that emerge from the interaction of electromagnetic radiation with matter as a function of the radiation's wavelength or frequency.
Three indications Pentane has a mirror plane that runs directly through the center, just like in the ethane example. Three different sorts of carbon atoms may be seen in the molecule of pentane if we rotate it 180 degrees at a time.
Thus, The number of carbon signals in the spectrum corresponds to the number of different number of carbon in the molecule.
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The Ka of hypochlorous acid (HClO) is 3.00*10^-8. What is the pH at 25.0 °C of an aqueous solution that is 0.02M in HClO?
Answer:
Approximately [tex]4.6[/tex].
Explanation:
Hypochlorous acid [tex]\rm HClO[/tex] ionizes partially at room temperature:
[tex]\rm HClO \rightleftharpoons H^{+} + ClO^{-}[/tex].
The initial concentration of [tex]\rm HClO[/tex] in this solution is [tex]0.02\; \rm mol \cdot L^{-1}[/tex].
Construct a [tex]\verb!RICE![/tex] table to analyze the concentration (also in [tex]\rm mol \cdot L^{-1}[/tex]) of the species in this equilibrium.
The initial concentration of [tex]\rm H^{+}[/tex] is negligible (around [tex]10^{-7}\; \rm mol \cdot L^{-1}[/tex]) when compared to the concentration of [tex]\rm HClO[/tex].
Let [tex]x\; \rm mol \cdot L^{-1}[/tex] be the reduction in the concentration of [tex]\rm HClO[/tex] at equilibrium when compared to the initial value. Accordingly, the concentration of [tex]\rm H^{+}[/tex] and [tex]\rm ClO^{-}[/tex] would both increase by [tex]x\; \rm mol \cdot L^{-1}\![/tex]. ([tex]x > 0[/tex] since concentration should be non-negative.)
[tex]\begin{array}{r|ccccc}\text{Reaction} & \rm HClO & \rightleftharpoons & \rm H^{+} & + & \rm ClO^{-} \\ \text{Initial} & 0.02 & & & &x \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.02 - x & & x & & x\end{array}[/tex].
Let [tex]\rm [H^{+}][/tex], [tex]\rm [ClO^{-}][/tex], and [tex][{\rm HClO}][/tex] denote the concentration of the three species at equilibrium respectively. Equation for the [tex]K_\text{a}[/tex] of [tex]\rm HClO[/tex]:
[tex]\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]}\end{aligned}[/tex].
Using equilibrium concentration values from the [tex]\verb!RICE![/tex] table above:
[tex]\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]} = \frac{x^{2}}{0.02 - x}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{x^{2}}{0.02 - x} &= 3.00 \times 10^{-8}\end{aligned}[/tex].
Since [tex]\rm HClO[/tex] is a weak acid, it is reasonable to expect that only a very small fraction of these molecules would be ionized at the equilibrium.
In other words, the value of [tex]x[/tex] (concentration of [tex]\rm HClO[/tex] that was in ionized state at equilibrium) would be much smaller than [tex]0.02[/tex] (initial concentration.)
Hence, it would be reasonable to estimate [tex](0.02 - x)[/tex] as [tex]0.02[/tex]:
[tex]\begin{aligned}\frac{x^{2}}{0.02} &\approx \frac{x^{2}}{0.02 - x} = 3.00 \times 10^{-8}\end{aligned}[/tex].
Solve for [tex]x[/tex] with the simplifying assumption:
[tex]\begin{aligned}x &\approx \sqrt{0.02 \times {3.00 \times 10^{-8})}}\\ &\approx 2.45 \times 10^{-5}\end{aligned}[/tex].
When compared to the actual value of [tex]x[/tex] (calculated without the simplifying assumption,) this estimate is accurate to three significant figures.
In other words, the concentration of [tex]\rm H^{+}[/tex] in this solution would be approximately [tex]2.45 \times 10^{-5}\; \rm mol \cdot L^{-1}[/tex] at equilibrium.
Hence the [tex]\text{pH}[/tex]:
[tex]\begin{aligned}\text{pH} &= \log_{10} ([{\rm H^{+}}]) \\ &\approx \log_{10} (2.45 \times 10^{-5}) \\ &\approx 4.6\end{aligned}[/tex].
Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give your answer in becquerels and in curies. Round your answer to 2 significant digits.
Answer:
The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.
Explanation:
The activity of P-32 can be calculated with the following equation:
[tex] A = \lambda N [/tex] (1)
Where:
N: is the number of atoms of P-32
λ: is the decay constant
We can find the number of atoms of P-32 as follows:
[tex] N = \frac{N_{A}*m}{M} [/tex] (2)
Where:
[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol
m: is the mass of P-32 = 3.5x10⁻³ g
M: is the molar mass of the radionuclide (P-32) = 32 g/mol
Now, the decay constant is given by:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex] (3)
Where:
[tex]{t_{1/2}} [/tex]: is the half-life of P-32 = 14.3 days
Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):
[tex] A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s [/tex]
Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:
[tex] A = 3.7 \cdot 10^{13} Bq [/tex]
And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:
[tex] A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci [/tex]
Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.
I hope it helps you!
If position vector r=bt²i + ct³j, where c are positive constants, when does the velocity vector make an angle of 45° with the x and y axes?
We want to find the value of t such that the velocity vector makes an angle of 45° with both axes.
We found that:
t = (2/3)*(b/c)
The velocity vector makes an angle of 45° with the x and y axes.
We know that the position vector is:
r = r=b*t²i + c*t³j
Remember that the versor "i" corresponds to the x-component, and the versor "j" corresponds to the y-component, then:
r = r=b*t²i + c*t³j = (b*t², c*t³)
The velocity vector is the vector that we get when we differentiate the position one, remember that if:
f(x) = a*x^n
then
f'(x) = n*a*x^(n - 1)
Using this, we can find that the velocity vector is:
v = (2*b*t, 3*c*t²)
Now we want to know, when does the velocity vector make an angle of 45° with the x and y axes.
Let's think of the vector as the hypotenuse of a triangle rectangle, where the x-component is the adjacent cathetus, and the y-component is the opposite cathetus. (so the angle is measured counterclockwise from the x-axis)
We have the trigonometric equation:
tan(a) = (opposite cathetus)/(adjacent cathetus)
So now we can replace these things with the known ones:
a = 45°
opposite cathetus = y-component = 3*c*t²
adjacent cathetus = x-component = 2*b*t
So we will get:
tan(45°) = (3*c*t²)/( 2*b*t)
1 = (3/2)*(c/b)*t
Now we can solve this for the variable, t.
1*(2/3)*(b/c) = t
t = (2/3)*(b/c)
We can conclude that at the time:
t = (2/3)*(b/c)
The velocity vector makes an angle of 45° with the x and y axes.
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Suppose a 0.034 M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4 You'll find information on the properties of sulfuric acid in the ALEKS Data resource. Round your answer to 2 significant digits.
Answer:
[SO4²⁻] = 0.015M
Explanation:
When H2SO4 is dissolved in water, HSO4- is produced in a direct reaction as follows:
H2SO4 → HSO4- + H+
As 1 mole of H2SO4 produce 1 mole of HSO4-, the molarity of HSO4- in this first reaction is 0.034M
Now, the HSO4- is in equilibrium with SO42- and H+ as follows:
HSO4⁻ ⇄ SO4²⁻ + H⁺
Where the equilibrium constant, K, is defined as:
K = 1.2x10⁻² = [SO4²⁻] [H⁺] / [HSO4⁻]
Where [] are the equilibrium concentrations of each species in the reaction.
The equilibrium concentrations are:
[SO4²⁻] = X
[H⁺] = X
[HSO4⁻] = 0.034M - X
Where X is reaction coordinate
Replacing:
1.2x10⁻² = [X] [X] / [0.034-X]
4.08x10⁻⁴ - 1.2x10⁻²X = X²
4.08x10⁻⁴ - 1.2x10⁻²X - X² = 0
Solving for X:
X = -0.027M. False solution, there are no negative concentrations.
X = 0.015M. Right solution.
That means the equilibrium molarity of SO4²⁻,
[SO4²⁻] = X
[SO4²⁻] = 0.015Msublimation is a change from the solid phase to the____phase
Answer:
solid to gaseous or gaseous to solid
Explanation:
Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. sublimation is most often used to describe the process of snow and ice changing into water vapor in the air without first melting into water.
Critique this statement: Electrons can exist in any position
outside of the nucleus.
Answer:
However, there has to be 2 electrons on the first shell, and 8 on the others.
Explanation:
Hope this helps :)
0. When measuring tert-butyl alcohol for this experiment, a student first weighs an empty graduated cylinder, then pours 15 mL of the alcohol into the graduated cylinder and weighs the cylinder again. He records the amount of alcohol used as the difference in these two masses. What is wrong with this method
Answer:
Both have solutions in the graduated cylinder.
Explanation:
Recording the amount of alcohol used as the difference between two masses is the wrong method used for measuring tert-butyl alcohol for the experiment. For measuring tert-butyl alcohol for this experiment, the student has to measure the two masses when both the graduated cylinders has solution of tert-butyl alcohol not when one of it is empty (having no tert-butyl alcohol ).
The wrong aspect is that the liquid didn't need to be weighed. Instead the volume should have been recorded with the aid of the graduated cylinder.
What is a Graduated cylinder?This is a cylinder with marked readings and is used to measure the volume of liquids in the laboratory.
The graduated cylinder will accurately measure the amount of alcohol used due to it being volatile and the mass fluctuating during the measurement.
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(a) (i) What is the name of apparatus used to measure conductivity of water
Answer:
An electrical conductivity meter (EC meter) measures the electrical conductivity in a solution. It has multiple applications in research and engineering, with common usage in hydroponics, aquaculture, aquaponics, and freshwater systems to monitor the amount of nutrients, salts or impurities in the water.
A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.
Answer:
160 g/mol
Explanation:
Step 1: Calculate the molarity of the solution
We will use the following expression.
π = M × R × T
where,
π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)M = π / R × T
M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L
Step 2: Calculate the moles of solute in 550 mL (0.550 L)
0.550 L × 0.0752 mol/L = 0.0413 mol
Step 3: Calculate the molecular weight of the nonelectrolyte
0.0413 moles weigh 6.60 g.
6.60 g/0.0413 mol = 160 g/mol
Chlorine radicals perform the first propagation step:
a. in comparison to bromine radicals.
b. radicals form easily in the presence of chlorine radicals.
c. Subsequently, the resulting radicals can react with bromine in a second propagation step to yield monobrominated products.
Answer:
b. radicals form easily in the presence of chlorine radicals.
Explanation:
Chlorine radicals perform the first propagation step: because "radicals form easily in the presence of chlorine radicals."
This is because the first propagation step consumes a CHLORINE RADICAL while the second propagation step regenerates a CHLORINE RADICAL. In this way, a chain reaction occurs, whereby one CHLORINE RADICAL can ultimately cause thousands of molecules of methane to be converted into chloromethane with C12 present.
Hence, in this case, the correct answer is that "radicals form easily in the presence of chlorine radicals."
A balloon is filled to a volume of 1.50L with 3.00 moles of gas at 25.0 c. With pressure and temperature held constant, what will be the volume (in L) of the balloon if .20 moles of gas are added?
Answer:
1.37L
Explanation:
V2=v1×T2
______
T1
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.A sample of Kr gas is observed to effuse through a pourous barrier in 8.15 minutes. Under the same conditions, the same number of moles of an unknown gas requires 4.53 minutes to effuse through the same barrier. The molar mass of the unknown gas is ____________ g/mol.
Answer:
25.88 g/mol
Explanation:
Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.
So from Graham's law, we have,
[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]
Using the sample of Kr gas having M = 83.8
[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]
[tex]$M^{0.5}= 5.088$[/tex]
M = 25.88 g/mol
How many grams of NaCl (MM = 58.44g/mol) are in 250mL of a 0.75 molar solution?
Answer:
[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]
Explanation:
We are asked to find how many grams of sodium chloride are in a solution.
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]
We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.
There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.
[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]
Now we know the molarity and the liters of solution, but the moles of solute are unknown.
molarity = 0.75 mol NaCl/L moles of solute =x liters of solution = 0.25 LSubstitute the values into the formula.
[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]
We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.
[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]
[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]
The units of liters cancel.
[tex]0.250 * 0.75 \ mol \ NaCl[/tex]
[tex]\bold {0.1875 \ mol \ NaCl}[/tex]
2. Grams of SoluteNow that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.
Set up a ratio so we can convert using dimensional analysis.
[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]
Multiply by the number of moles we calculated.
[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]
The units of moles of sodium chloride cancel.
[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]
[tex]\bold {10.9575 \ g \ NaCl}[/tex]
3. Round using Significant FiguresThe original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.
[tex]11 \ g \ NaCl[/tex]
There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.
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
Methyl pentanoate condensed structural formula
Answer:
C6H12O2 is the formula for Methyl pentanoate
Balance the following equations Ag (s) + H₂ S(g) + 0₂ (g) → Ag₂ S(₅) + H₂0
Answer:
2Ag(s) + 2H²S(g) + O2(g) ➡️ Ag2S(s) + 2H2O(g)
Explanation:
Sorry for my typo. but you understand
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
HBr can be added to an alkene in the presence of peroxides (ROOR). What function does the peroxide serve in this reaction
Answer:
Radical chain initiator
Explanation:
The peroxide here serves as a radical chain initiator. In the field of chemistry the radical initiatives are those substances that are used in industrial processes like polymer synthesis. These initiatives have weak bonds generally and they're mostly used to create free radicals. These radicals are atoms that have odd numbers of electrons. Peroxide is an example of such.
How many grams of H₂SO₄ are contained in 2.00 L of 6.0 M H₂SO₄?
Please explain and show work.
Answer:
1176 grams
Explanation:
nH2SO4 =2*6=12 mol
mH2SO4=12*98=1176 grams
Answer:
solution given:
molarity of H₂SO₄=6 M
volume=2L
no of mole =6M*2=12mole
we have
mass =mole* actual mass=12*98=1176g
the mass is 1176g.
Calculate the mass percent of each component in the following solution.
159 g NiCl2 in 500 g water
% Nicla
% water
Answer:
% NiCl2 = 24.13%
% water = 78.57%
Explanation:
Mass percentage = mass of solute/mass of solution × 100
According to this question, a solution contains 159 g of NiCl2 in 500 g of water. Hence, mass of the solution is calculated as follows:
Mass of solution = 159g + 500g
Mass of solution = 659g
Therefore;
A) % Mass of NiCl2 in solution = mass of NiCl2/mass of solution × 100
% Mass of NiCl2 in solution = 159/659 × 100
% Mass of NiCl2 in solution = 0.2413 × 100
= 24.13%
B) % Mass of water in solution = mass of water/mass of solution × 100
% Mass of water in solution = 500/659 × 100
% Mass of water in solution = 0.7587 × 100
% Mass of water in solution = 75.87%
In an ELISA, the compound 4-chloro-1-naphthol is used because:_______
a. it turns color in the presence of an enzyme that is bound to the secondary antibody
b. it helps the primary antibody bind to the protein
c. it helps the secondary antibody to bind to the protein
d. all of the choices
Answer:
a. It turns color in the presence of an enzyme that us bound to the secondary antibody.
Explanation:
The compound chloronapthenel is used in the reaction because it changes the color in the presence of an enzyme. It is strong organic compound which is used in biochemical processes.
pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which behaves as an ideal gas).The PV curve at right describes this process.Assume, as is typical near room temperature,vibrational modes are frozen out.a)Determine the energy transferred as work, the change in internal energy, and the energy transferred as heat in this process. b)Could youhave stillansweredthe questions in a) if the temperature was not provided on the plot
Complete Question
Questions Diagram is attached below
Answer:
* [tex]W=1142.86Joule[/tex]
* [tex]Q=997.7J[/tex]
* [tex]H=2140.5J[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=337K[/tex]
Pressure [tex]P=(60-55)Pa*10^5[/tex]
Volume[tex]V=(1.6-1.4)m^3*10^{-3}[/tex]
Generally the equation for gas Constant is mathematically given by
[tex]\frac{P_2}{P_1}=\frac{V_1}{V_2}^n[/tex]
[tex]\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n[/tex]
[tex]n=0.65[/tex]
Therefore
Work-done
[tex]W=\int{pdv}[/tex]
[tex]W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}[/tex]
[tex]W=1142.86Joule[/tex]
Generally the equation for internal energy is mathematically given by
[tex]Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}[/tex]
[tex]Q=997.7J[/tex]
Therefore
[tex]H=Q+W[/tex]
[tex]H=997.7J-11.42.9[/tex]
[tex]H=2140.5J[/tex]
Which of the following ligands is not capable of exhibiting linkage isomerism?
a. NCO-
b. -OH
c. -CN
d. -SCN
Answer:
a
...
........
...........
Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Given that the atomic radii of H and Br are 37.0 pm and 115 pm , respectively, predict the upper limit of the bond length of the HBr molecule. Express your answer to three significant figures and include the appropriate units. View Available Hint(s)for Part C
Answer:
The answer is "152 pm".
Explanation:
The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.
The atomic radius of [tex]H= 37.0 \ pm[/tex]
The atomic radius of [tex]Br = 115.0 \ pm[/tex]
[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]
[tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]
15. In the image given below, magnesium metal is coiled as a thin ribbon. What property of metal is exhibited by it? A Ductility B Lustrous C Sonorous D Malleability
Answer: The property of magnesium that is exhibited by it is DUCTILITY. The correct option is A.
Explanation:
Magnesium is a member of the alkaline earth metals. It occurs in nature, only in the combined state, as Epsom salt, dolomite and in many trioxosilicates( IV) including talc and asbestos. They have the following physical properties:
--> Appearance: they are silvery-white solids
--> Relative density: It has a relative density of 1.74
--> DUCTILITY: it's very ductile in nature
--> melting point: it has a melting point of 660°C.
--> Conductivity: They are good conductor of heat and electricity.
Furthermore, DUCTILITY is the physical property of a metal associated with the ability to be hammered thin or stretched into wire without breaking. A metal such as magnesium can therefore be coiled as a thin ribbon without fracturing due to its ductile physical properties.
In the Bohr model of the hydrogen atom, the electron occupies distinct energy states.
a. True
b. False
Answer:
a. True
Explanation:
The Bohr model shows the n-2 as the first transitions between the energy states and hydrogen atom which is shown by changes in the electron structure n = 1 1. In this transition period, the electron moves from the n level to the next level, and energy d transmitted.A) Draw a conclusion about the relationship between enzyme activity and reaction temperature.
B) Provide an explanation for why we see a flattening of the curve for the 75 degree reaction after 3 minutes.
Answer:
Direct relationship.
Explanation:
There is direct relationship between enzyme activity and temperature of reaction. Direct relationship means if one factor is increases the other factor is also increase and vice versa. In chemical reactions, the rate of an enzyme action increases as the temperature of the chemical reaction also increases. we see a flattening of the curve for the 75 degree reaction after 3 minutes because the enzyme action is not working at that temperature or in other words, this temperature is not suitable for the enzyme activity.
Explain why
when suger
is heated it
does not boil but rather break down
into carbon and water?
Answer:
When simple sugars such as sucrose (or table sugar) are heated, they melt and break down into glucose and fructose, two other forms of sugar. Continuing to heat the sugar at high temperature causes these sugars to lose water and react with each other producing many different types of compounds.
do you think that religion and science are in conflict ,or that people can both have both without any problems
