Answer:
d. Pergoldic acid
Explanation:
chlorate = ClO₃⁻
Goldate GoO₃⁻
On this basis we can infer that HClO₄ and HGoO₄ will be similar .
HClO₄ is called perchloric acid , hence HGoO₄ will be called pergoldic acid.
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?
Answer:
3.13%.
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 0.15
Half life (t½) = 2 mins
Number of half-life (n) = 5
Fraction of sample remaining =.?
Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:
Amount remaining (N) = 1/2ⁿ × original amount (N₀)
NOTE: n is the number of half-life.
N = 1/2ⁿ × N₀
N = 1/2⁵ × 0.15
N = 1/32 × 0.15
N = 0.15/32
N = 4.69×10¯³
Therefore, 4.69×10¯³ is remaining after 5 half-life.
Finally, we shall the fraction of the sample remaining after 5 half-life as follow:
Original amount (N₀) = 0.15
Amount remaining (N) = 4.69×10¯³
Fraction remaining = N/N₀ × 100
Fraction remaining = 4.69×10¯³/0.15 × 100
Fraction remaining = 3.13%
What concentration of NO−3NO3− results when 897 mL897 mL of 0.497 M NaNO30.497 M NaNO3 is mixed with 813 mL813 mL of 0.341 M Ca(NO3)2?
Answer:
Explanation:
NaNO₃ = Na⁺ + NO₃⁻¹
.497 M .497 M
moles of NO₃⁻¹ = .897 x .497 = .4458 moles
Ca( NO₃)₂ = Ca + 2 NO₃⁻¹
.341 M 2 x .341 M = .682 M
moles of NO₃⁻¹ = .813 x .682 = .5544 moles
Total moles = .4458 moles + .5544 moles
= 1.0002 moles
volume of solution = 897 + 813 = 1710 mL
= 1.710 L
concentration of nitrate ion = 1.0002 / 1.710 M
= .585 M
what might happen to variables in a science experiment that would lead to unusable results?
Answer:
ejeb094
Explanation:
nnb3neneie9eei rje
SOMEONE PLZ HELP!!!!
Answer:
4.22mL
Explanation:
V=m/d
v= 18.45g/4.37g/mL
Bromine is found above iodine in Group 17 of the periodic table. If an ion formed by bromine has a charge of 1-, what is the charge on an ion
formed by iodine?
A. -7
B. -2
C. -1
D. +1
Answer:
Explanation:
Iodine is the most flexible of the atoms in Group 17 of the periodic table. It can have more than 1 charge, especially when combined with oxygen. The answer that you are expected to give, I think is C. Iodine's dominant charge is - 1
A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to produce carbon dioxide CO2 and water H2O. What mass of carbon dioxide is produced by the reaction of 7.58 g of octane? Please explain the answer to me like I'm five, I want to understand but the content makes no sense.
Answer:
Mass = 23.232 g
Explanation:
Given data:
Mass of C₈H₁₈ = 7.58 g
Mass of CO₂ produced = ?
Solution:
Chemical equation:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Number of moles of octane:
Number of moles = mass/molar mass
Number of moles = 7.58 g/ 114.23 g/mol
Number of moles = 0.066 mol
Now we will compare the moles of CO₂ with octane from balance chemical equation.
C₈H₁₈ : CO₂
2 : 16
0.066 : 16/2×0.066 = 0.528
Mass of CO₂ produced:
Mass = number of moles × molar mass
Mass = 0.528 mol × 44 g/mol
Mass = 23.232 g
3. A certain Chemical Industry company has a quality control job opening. The job is open for any major with basic knowledge of chemistry. You decided to apply. In the interview the HR personnel gives you a sealed folder from a certain lot to test your laboratory experience, as well as your quantitative and volumetric analysis skills. The chemical contained in the sample is benzoic acid (C-H602) and it is known to be a monoprotic acid. In order to get the job, you need to determine if the sample's purity is acceptable based on their standards. Inside the folder you found a vial with a solid sample labeled BA-I, a periodic table, and the following data: 1.250 g of the sample required 20.15 mL of 0.500 M concentration of NaOH to reach the end point. The lot can be denied if the purity is below 99.5 % purity.
1) What is the purity in the sample?
2) Is it the purity acceptable?
3) Would you repeat the titration experiment?
Answer:
1) 97.6%
2) No the purity is not acceptable because the standard is 99.5% purity.
3) Yes I will repeat the titration experiment to confirm my result.
Explanation:
Equation of the reaction;
C7H6O2(aq) + NaOH(aq) ---------> C7H5ONa(aq) + H2O(aq)
From the information provided;
Number of moles of NaOH reacted = concentration × volume = 20.15/1000 × 0.500 = 0.01 moles
From the reaction equation;
1 mole of C7H6O2 reacts with 1 mole of NaOH
Hence 0.01 moles of C7H6O2 will react with 0.01 moles of NaOH
Mass of C7H6O2 reacted = number of moles of C7H6O2 × molar mass of C7H6O2
Molar mass of C7H6O2 = 122.12 g/mol
Mass of C7H6O2 reacted = 0.01 moles × 122.12 g/mol = 1.22 g
Percentage by mass of pure C7H6O2 in the impure sample = 1.22/1.250 × 100 = 97.6 %
What is the momentum of a 1kg ball moving at 5m/s?
Answer:
5Ns
momentum= mass *velocity
=1*5
=5Ns
The Adálie penguins feed on krill that live on the underside of ice sheets. Which of the following would help Adálie penguins survive during the melting of Antarctic sea ice?
The correct answer is Migrating farther into the sea
Explanation:
The Adálie penguins rely on ice sheets to obtain food as the krill they feed on is only found underside of the ice. This implies as the Antarctic sea ice melts Adálie penguins need to look for new places to find krill as if there are no ice sheets there are no krill for these penguins. Due to this, it is likely these penguins are forced to migrate or move farther into the sea to survive because it is likely in these zones there are ice sheets and therefore places where they can find krill. Despite this, this process might lead to a significant decline in the population of penguins due to the competence for food and the process of migration itself. Thus, the one that would help Adáline penguins to survive during the melting of Antarctic sea ice is migrating farther into the sea.
Have a nice day! :)
The Adálie penguins feed on krill that live on the underside of ice sheets. Adálie penguins survive during the melting of Antarctic sea ice help migrating farther into the sea. Therefore, option A is correct.
What is migration ?The term migration is defined as the movement of either people or animals from one area to another.
The Adálie penguins rely on ice sheets for food because the krill they eat is only found on the underside of the ice. As a result, it is likely that these penguins are forced to migrate or move further into the sea to survive.
Regardless, this process may result in a significant decline in penguin population due to food competence and the migration process itself.Thus, the one that will assist Adáline penguins in surviving during the melting of Antarctic sea ice is migrating deeper into the sea.
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Your question is incomplete,most probably your question was
The Adálie penguins feed on krill that live on the underside of ice sheets. Which of the following would help Adálie penguins survive during the melting of Antarctic sea ice? Migrating farther into the sea Breeding during summer months Growing strong toe nails to grip ice Spending more time raising their young
What happen when a piece of sodium is exposed in air?
Answer:
I literally got this from google.
Explanation:
In ordinary air, sodium metal reacts to form a sodium hydroxide film, which can rapidly absorb carbon dioxide from the air, forming sodium bicarbonate. ... In a comparatively dry atmosphere, sodium burns quietly, giving off a dense white caustic smoke, which can cause choking and coughing.
Happy almost Halloween! :)
The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga
Answer:
2.35 x 10²⁰ atoms Ga
Explanation:
After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.
[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]
Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.
The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
StoichiometryFrom the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.
First, we will determine the number of moles of Ga present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]
Mass = 27.2 mg = 0.0272 g
Molar mass = 69.72 g/mol
Then,
[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]
[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles
Now, for the number of atoms present
From the formula
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of Ga atoms = 0.000390132 × 6.022×10²³
Number of Ga atoms = 2.35 × 10²⁰ atoms
Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
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plz help answer both will mark brainest
3. Which of the following molecules would want except to have a nonpolar covalent bond
Answer:
polar bonds are caused by different kind of atoms, because almost every atoms have different powers to attract electrons.
the answer will be the two same atoms, F2
Is nuclear fission exothermic or endothermic? Explain your answer.
Answer:
Exothermic
Explanation:
Nuclear fission means splitting, so there is a lot of energy being released.
Use the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.
Answer:
The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].
The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].
Explanation:
The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.
Let [tex]\lambda_\text{vac}[/tex] denote the wavelength of the photon released when measured in vacuum.Let [tex]R_\text{H}[/tex] denote the Rydberg constant for hydrogen. [tex]R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}[/tex].Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the principal quantum number of the initial and final main energy level of that electron. (Both [tex]n_1\![/tex] and [tex]n_2\![/tex] should be positive integers; [tex]n_1 > n_2[/tex].)The Rydberg Equation gives the following relation:
[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].
Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:
[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].
In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:
[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].
Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].
Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:
[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].
Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].
Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:
[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].
The energy of the transition between n = 7 and n = 3 is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.
Using the Rydberg Equation for energy;
ΔE = -RH(1/n^2final - 1/n^2initial)
Given that;
nfinal = 3
ninitial = 7
RH = 2.18 × 10^-18 J
ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)
ΔE = - 2.18 × 10^-18(0.11 - 0.02)
ΔE = - 1.96 × 10^-19 J
For the second part;
Since the wavelength is 1000nm, we have;
λ = 1000nm
c = 3 × 10^8 m/s
f = ?
c = λf
f = c/λ
f = 3 × 10^8 m/s/1000 × 10^-9 m
f = 3 × 10^8 m/s/ 1 × 10^-6 m
f = 3 × 10^14 Hz
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pentane or 2,2,3- trimethylhexane has the higher boiling point. why?
Answer:
2,2,3- trimethylhexane because it has more carbon atoms than pentane.
Explanation:
Hello.
In this case, since the physical properties of organic compounds are intensified as the number of carbon atoms start increasing on it, the larger the amount of carbon atom, the higher the boiling point since more energy is required to allow the liquid-phase molecules to transcend to the vapor-phase.
In such a way, since pentane has five carbon atoms and 2,2,3- trimethylhexane has nine carbon atoms, 2,2,3- trimethylhexane has the highest boiling point.
Best regards.
1. Assume this experiment (after the extractions are complete) left you with 4 pure products (aspirin, acetaminophen and caffeine, and the binder). Your lab mate distracted you after you had labeled the binder and now you can’t remember which of the remaining 3 products is which. What would be a simple test that you can perform in the laboratory to distinguish between the 3 remaining solids isolated from your experiments?
Answer:
Explanation:
Out of aspirin, acetaminophen and caffeine, aspirin is an acid because it is acetyl saliciylic acid . Hence it can be tested with litmus paper .
acetaminophen contains phenolic functional group , hence it is a weak acid . It can be tested with any test with which phenol test are done, like with neutral solution of ferric chloride .
caffeine is weak basic substance . It can also be tested with the help of testing a basic substance .
20 POINTS PLEASE ANSWER ASAP!!!!
Why is calcium (Ca) in group 2, period 4 on the periodic table?
A Calcium, like all group 2 elements, is nonreactive and a gas at room temperature.
B Calcium, like all period 4 elements, is nonreactive and a gas at room temperature,
C Calcium, like all group 2 elements, is reactive and a solid at room temperature.
D. Calcium, like all period 4 elements, is reactive and a solid at room temperature.
Calcium (Ca) is in group 2 and period 4 on the periodic table be because Calcium has 2 valence electrons and 4 electron shell. Thus, calcium is a metal like all other group 2 element.
The correct answer to the question is Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.
Calcium is a group 2 element majorly because it has 2 valence electrons. It is also in period 4 because it has 4 electron shells.
Being a group 2 element, calcium is a solid at room temperature and also reactive. All elements in the group 2 are metals.
There are other elements in period 4 which are not solid. For example krypton is an element in period 4 and it is a gas and not reactive.
From the above information, we can conclude that the correct answer to the question is:
Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.
See attached image
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Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6% Write the empirical chemical formula of X.
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Answer:
The empirical formula of X is C3H4O3.
Explanation:
Peroxyacylnitrate (PAN) is one of the components
of smog. It is a compound of C, H, N, and O.
Determine the percent composition of oxygen and
the empirical formula from the following percent
composition by mass: 19.8 percent C,
2.50 percent
H, 11.6 percent N. What is its molecular formula
given that its molar mass is about 120 g?
C – 19,9%, H – 2,2%, N – 11,6%, O – x%
[tex]M=120\frac{g}{mol}[/tex]
1 percentage
The entire molecule is 100% and all the components of the compound add up to 100%.
100% - 19,9% - 2,5% - 11,6% = 66,1%
The compound contains 66,1% oxygen.
2 molar masses
[tex]M_{C}=12,01\frac{g}{mol}[/tex]
[tex]M_{H}=1,008\frac{g}{mol}[/tex]
[tex]M_{O}=15,999\frac{g}{mol}[/tex]
[tex]M_{N}=14,007\frac{g}{mol}[/tex]
3 masses
The compound has a molar mass of 120g/mol. So one molecule weighs 120 g. To find out how much the percentage of a component weighs, you have to calculate it using the molar mass.
carbon
19,8% of 120g
[tex]m=120g*0,198\\m=23,76g[/tex]
One molecule contains 23,76g of carbon.
hydrogen
2,5% of 120g
[tex]m=120g*0,025\\m=3g[/tex]
One molecule contains 3g of hydrogen.
oxygen
66,1% of 120g
[tex]m=120g*0,661\\m=79,32g[/tex]
One molecule contains 79,32g of oxygen.
nitrogen
11,6% of 120g
[tex]m=120g*0,0,116\\m=13,92g[/tex]
One molecule contains 13,92g of nitrogen.
4 amount of substance
carbon
[tex]n=\frac{23,76g}{12,01\frac{g}{mol} }\\n=1,98mol[/tex]
The compound contains about 2 moles of carbon.
hydrogen
[tex]n=\frac{3g}{1\frac{g}{mol} }\\n=3mol[/tex]
The compound contains about 3 moles of hydrogen.
oxygen
[tex]n=\frac{79,32g}{15,999\frac{g}{mol} }\\n=4,96mol[/tex]
The compound contains about 5 moles of oxygen.
nitrogen
[tex]n=\frac{13,92g}{14,007\frac{g}{mol} }\\n=0,99mol[/tex]
The compound contains about 1 moles of nitrogen.
5. molecular formula
The formula results from the ratio of the amounts of substance.
[tex]n_{C} :n_{H} :n_{O} :n_{N} =2:3:5:1\\C_{2}H_{3}NO_{5}[/tex]
The molecular formula of the given compound is C₂H₃NO₅, and percent composition of oxygen in it is 66.1%.
How do we calculate mass from % composition?Mass of any composition of any compound will be calculated by using the below formula as:
Mass of component = (% composition)×(mass of compound) / 100
Given mass of compound = 120g/mol
Total composition of compound (100%) = Percent composition of all components
% composition of oxygen = 100 - (19.8 + 2.50 + 11.6) = 66.1%
Moles will be calculated as:
n = W/M, where
W = given mass
M = molar mass
For carbon atom:Mass of Carbon component = (0.198)(120g) = 23.76g
Moles of Carbon atom = 23.76g / 12.01g/mol = 1.98mol = 2 moles
For nitrogen atom:Mass of Nitrogen component = (0.116)(120g) = 13.92g
Moles of Nitrogen atom = 13.92g / 14.007g/mol = 0.99mol = 1 moles
For oxygen atom:Mass of Oxygen component = (0.661)(120g) = 79.32g
Moles of Oxygen atom = 79.32g / 15.99g/mol = 4.96mol = 5 moles
For hydrogen atom:Mass of Hydrogen component = (0.025)(120g) = 3g
Moles of Hydrogen atom = 3g / 1g/mol = 3 moles
So, the molecular formula of the compound on the basis of moles of given entities is C₂H₃NO₅.
Hence required molecular formula is C₂H₃NO₅.
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Which is one way that minerals crystallize from materials dissolved in water?
from the air
from solutions that evaporate
from hot water solutions when water boils
from the soil
Answer:
the second answer its science behind it
Answer:
b
Explanation:
Question 5 of 5
Which two phrases describe the nature of an electromagnetic force?
O A. Acts only when objects touch each other
B. Produced by interactions between magnetic objects
O c. Not a fundamental force of nature
O D. Produced by interactions between electrically charged objects
Answer:
I think it's A and D
Explanation:
I'm not sure if it's right
Answer:
The answer is B and D
Explanation:
trust fr
3. Does matter increase in
mass when mixed with
another substance?
a Yes
b. No
Answer:
b. No
Explanation:
The question above is related to "The Law of Conservation of Mass." This law states that mass is neither created nor can it be destroyed. Even if a particular matter will be mixed with another substance or it undergoes chemical reaction, the mass of the matter will remain the same. Even with a change in phase, the mass of the matter will remain the same.
PLS PLS PLS PLS PLS PLS PLS help ASAP!!!!! Scientists call all of the compounds that contain carbon and are found in living things Organic because ________.
WILL DO BRAINLIEST!
2. Determine the possible traits of the calves of : Da red (RR) bull is mated with a red (RR) cow 1 a red * (RR) bullis mated with a white (WW) Cow 2 Da roan * (RW) is mated with a red(RR)Cow 3 3. Illustrate your answers using a Punnett square. 4. Write your answers on the paper.
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic acid has a pKa of 3.75. 2. Create approximately 20 mL of buffer solution with a pH of 4.25.
Answer:
15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.
Explanation:
To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid
4.25 = 3.75 + log [A⁻] / [HA]
0.5 = log [A⁻] / [HA]
3.162 = [A⁻] / [HA] (1)
As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:
0.10M * 20x10⁻³L =
2x10⁻³moles = [A⁻] + [HA] (2)
Replacing (2) in (1):
3.162 = 2x10⁻³moles - [HA] / [HA]
3.162 [HA] = 2x10⁻³moles - [HA]
4.162[HA] = 2x10⁻³moles
[HA] = 4.805x10⁻⁴ moles
[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles
That means, to create the buffer you must add:
[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =
15.2mL of the 0.10M sodium formate solution[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =
4.8mL of the 0.10M formic acid solutionNicotinic acid, HC6H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4 × 10-5. Calculate [H+] and the pH of a 0.041 M solution of HC6H4NO2.
Answer:
[H+] = 7.576x10⁻⁴M
pH = 3.12
Explanation:
Based on the equilibrium of the nicotinic acid in water:
HC6H4NO2(aq) + H2O(l) ⇄ C6H4NO2-(aq) + H3O+(aq)
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
As both C6H4NO2-(aq) and H3O+(aq) comes from the same equilibrium, we can approximate their concentration as X and replace:
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
1.4x10⁻⁵ = [X] [X] / [0.041M]
5.47x10⁻⁷ = X²
7.576x10⁻⁴M = X = [H+]And as pH is defined as -log [H⁺]
pH = 3.12A gas has a volume of 300 mL and a pressure of 2 atm. What volume will the gas occupy when the pressure is
increased to 7 atm (total)?
Answer:
The answer is 85.71 mLExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
We have
[tex]V_2 = \frac{300 \times 2}{7} = \frac{600}{7} \\ = 85.714285...[/tex]
We have the final answer as
85.71 mLHope this helps you
Consider the diagram below.
What does C represent?
A) enthalpy of reaction
B) activation energy
C) activated complex
D) energy of the reactants
Answer:
A) enthalpy of reaction
Explanation:
The region C signifies the enthalpy of reaction.
This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.
Enthalpy changes are heat changes accompanying a physical and chemical change. An enthalpy is the difference between the sum of the heat contents of products and sum of the heat contents of reactants.it is indeed A) enthalpy of reaction
ou are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution
Complete Question
You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the enantiomeric excess?
Answer:
The value is [tex]k = 96 %[/tex]
Explanation:
From the question we are told that
The percentage of enantiomer A is A = 98%
The percentage of enantiomer B is B = 2%
Generally the enantiomeric excess is mathematically represented as
[tex]k = \frac{A -B}{A+B} * 100[/tex]
=> [tex]k = \frac{98 -2}{98+2} * 100[/tex]
=> [tex]k = 96 %[/tex]