Final answer:
The epicenter of an earthquake can be located using the arrival times of S- and P-waves at three or more seismographic stations, allowing for triangulation and precise determination of the distance to the earthquake source.
Explanation:
To locate the epicenter of an earthquake, seismographic stations use the arrival times of S- and P-waves. The S-wave is slower than the P-wave, so the difference in arrival times at three or more seismographic stations can be used to determine the distance to the epicenter. By triangulating the distances from multiple stations, the epicenter can be pinpointed.
For example, if the S-wave and P-wave travel at speeds of 4.00 km/s and 7.20 km/s respectively, the difference in arrival times can be used to calculate the distance. The precision of the seismographs in measuring the arrival times allows for a precise determination of the distance to the earthquake source.
It is important to note that uncertainties in the propagation speeds of the waves can introduce greater uncertainty in determining the distance to the epicenter.
A match has about 21 milligrams of red phosphorus coating the tip. How many atoms of phosphorus is this?
Answer:
4.083 * 10^20 atoms.
Explanation:
One Mole of phosphorus contains 6.022 * 10^23 atoms (Avogadros number)'
Since 1 mole of Phosphorus has a mass of 30.974 grams, 21 milligrams has
6.022 * 10^23 * 0.021 / 30.974
= 0.004083 * 10^23
= 4.083 * 10^20
Which is usually the slowest way to regulate a reaction in a metabolic pathway?
A) Covalent modification
B) Changing the enzyme concentration
C) Allosteric modulation
All of the above are usually equally as fast.
Answer:
B) Changing the enzyme concentration
The slowest way to regulate a reaction in a metabolic pathway is by changing the enzyme concentration.
Metabolic Pathway:It is series of chemical reactions that occurred in the cell. Metabolites referred to as species participating in an enzymatic reaction.
Role of Enzymes:
Enzymes act as catalysts – they allow a reaction to proceed more rapidly – and they also allow the regulation of the rate of a metabolic reaction
The binding of an enzyme to its substrate lowers the activation energy of the reaction. If an enzyme is present, the amount of energy needed to make a product is lowered.
In metabolic reactions there is transfer of electrons from one compound to another by processes catalyzed by enzymes.
Enzymes participates in increasing and decreasing the metabolism process inside our body. On increasing the substrate concentration, the rate of enzyme activity also increases.
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Which correctly lists the following in order of increasing entropy? i. 1 mol of HCl (g) at 50 ° C ii. 1 mol of NaCl (s) at 25 ° C iii. 2 mol of HCl (g) at 50 ° C iv. 1 mol of HCl (g) at 25 ° C
Answer:
The order of increasing entropy will be:
1 mol of NaCl (s) at 25 ° C < 1 mol of HCl (g) at 25 ° C< 1 mol of HCl (g) at 50 ° C < 2 mol of HCl (g) at 50 ° C
Explanation:
The entropy increases with
a) increase in temperature
b) state of matter (the entropy order of matter is gas > liquid > solid)
So here
i) 1 mol of HCl (g) at 50 ° C : it is gas at high temperature as compared to HCl gas at lower temperature.
ii. 1 mol of NaCl (s) at 25 ° C : this is solid so will have lower entropy than gas.
iii. 2 mol of HCl (g) at 50 ° C: the moles are more so will have more entropy than 1 mol of HCl (g) at 50 ° C
iv. 1 mol of HCl (g) at 25 ° C : will have lower entropy than HCl gas at higher temperature but will have higher entropy than solid NaCl.
The correct order of increasing entropy is given as:
1 mol of NaCl (s) at 25 ° C < mol of HCl (g) at 50 ° C < 1 mol of HCl (g) at 25 ° C < 2 mol of HCl (g) at 50 ° C
What is entropy?Entropy can be defined as that measure of thermal energy of a system per unit temperature that is unavailable for doing work.
Mathematically, the entropy change in a chemical reaction is given by:
The sum of the entropies of the products - the sum of the entropies of the reactants.
In conclusion, the correct order of increasing entropy is given as
1 mol of NaCl (s) at 25 ° C < mol of HCl (g) at 50 ° C < 1 mol of HCl (g) at 25 ° C < 2 mol of HCl (g) at 50 ° C
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Which of the following accurately characterizes the process of freezing?
Because energy is released, freezing is an exothermic process.
Because energy is absorbed, freezing is an exothermic process.
Because energy is released, freezing is an endothermic process.
Because energy is absorbed, freezing is an endothermic process.
Answer:
because energy is released
Explanation:
freesing is an exothermic process
C. Because energy is released, freezing is an endothermic process.
what is the endothermic process?In thermochemistry, an endothermic system is any process with an increase in the enthalpy H of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transferred into the system.
Endothermic reactions are chemical reactions wherein the reactants absorb heat energy from the surroundings to form products. these reactions lower the temperature in their surrounding region, thereby creating a cooling effect.
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Atoms- 1 2 3 4
Protons- 8 10 9 8
Neutrons- 10 9 9 11
Electrons- 8 10 9 8
Which two are isotopes of one another?
Answer:
The ones with 8 protons
Explanation:
Since there are two of them with 8 protons, we can assume they are the same element. The first 8 proton element has 10 neutrons while the second has 11. This makes them isotopes of one another
After an experiment is complete and has been published, what process makes sure the results are reliable?
A. Falsifiability
B. Peer review
C. The conclusion
D. Error Bars
Answer:
B. Peer review
Explanation:
Peer review ensures that the results of an experimental procedures are consistent are reliable and they meet their objective statement.
When peers which are professionals in a field of study subjects the results from an experiment into a test, they can give their own verdict as to wether such findings are consistent and reliable with the problem in view.
Answer:
the answer would be peer review
Explanation:
founders education chemistry
Calculate the heat of decomposition for this process at constant pressure and 25°C: CaCO3(s) → CaO(s) + CO2(g) The standard enthalpies of formation for the reactant are: ΔHf CaCO3(s) = −1206.9 kJ/mol ; ΔHf CaO(s) = −635.6 kJ/mol; ΔHf CO2(g) −393.5 kJ/mol
Answer:
177.8kJ/mol
Explanation:
In this reaction, the heat of decomposition is the same as the heat of formation. This is a decomposition reaction.
Given parameters:
ΔHf CaCO₃ = -1206.9kJ/mol
ΔHf CaO = −635.6 kJ/mol
ΔHf CO₂ = −393.5 kJ/mol
The heat of decomposition =
Sum of ΔHf of products - Sum of ΔHf of reactants
The equation of the reaction is shown below:
CaCO₃ → CaO + CO₂
The heat of decomposition = [ -635.6 + (-393.5)] - [−1206.9 ]
= -1029.1 + 1206.9
= 177.8kJ/mol
The heat of decomposition given by the difference in the ΔH of product and reactant is 177.8 kj/mol
Given the equation for the decomposition process :
CaCO3(s) → CaO(s) + CO2(g)The heat for the decomposition process is the difference of the sum of heat of the product and the heat of the reactant :
Reactant :
CaCO3(s) = - 1206.9 kj/molProduct :
CaO(s) + CO2(g) = - 635. 6 + (-393.5) = - 1029.1 kj/molHeat of decomposition :
ΔHproduct - ΔHreactantHeat of decomposition = - 1029.1 - (- 1206.9)
Heat of decomposition = -1029.1 + 1206.9 = 177.8 kj/mol
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The initial explanation for a phenomenon prior to experimentation is called a:
theory
law
hypothesis
rule
Answer: Hypothesis
Explanation: Hypothesis is a kind of idea that has been put forward prior to experimentation and whose results are not out yet.
The laws makes the theory then there are some rules which are needed to be followed in order to make the most out of the experiments. Then prior to the experiments , there comes the hypothesis whis the explanation of the work and which would be tested through the experiments.
Why might an idea or hypothesis be discarded? A. If another scientist doesn't like it. B. If evidence also supports another hypothesis. C. If it is tested and the evidence does not support it. D. If a politician disagrees with it.
Answer:
C. If it is tested and the evidence does not support it.
Explanation:
A hypothesis is more less a scientific guess. Before such a guess or prediction is made, empirical observations and deductions are first made. It is from the result of the observations that a hypothesis statement is made.
For a hypothesis to become widely adopted and accepted, it must be testable within the limits of the experiment as described by the proposer. When subjected to test and it agrees, the status of a hypothesis can be upgraded.
If the hypothesis is tested and evidence contrasts the result being sort for, a hypothesis will be discarded.
An _____ reaction results when the energy released by the formation of products is greater than the energy required to break the bonds in the reactants.
Answer:
Exothermic
Explanation:
An exothermic reaction results when the energy released by the formation of products is greater than the energy required to break the bonds in the reactants.
Calculate the standard enthalpy of formation of carbon disulfide (CS2) from it's elements, given that C(graphite) + O2(g) → CO2(g) ΔH o rxn = −393.5 kJ/mol S(rhombic) + O2(g) → SO2(g) ΔH o rxn = −296.4 kJ/mol CS2 + 3O2(g) → CO2(g) + 2SO2(g) ΔH o rxn = −1073.6 kJ/mol
To calculate the standard enthalpy of formation of carbon disulfide (CS2), we can use Hess's Law and the given enthalpy changes for the reactions involving carbon, sulfur, and oxygen. The standard enthalpy of formation of CS2 from its elements is -310.3 kJ/mol.
Explanation:To calculate the standard enthalpy of formation of carbon disulfide (CS2) from its elements, we can use Hess's Law and the given enthalpy changes for the reactions involving carbon (C), sulfur (S), and oxygen (O).
First, we can use the given reaction for the formation of carbon dioxide (CO2) to find the enthalpy change for the reaction involving carbon:
C(graphite) + O2(g) → CO2(g) ΔH o rxn = −393.5 kJ/mol
Next, we can use the given reaction for the formation of sulfur dioxide (SO2) to find the enthalpy change for the reaction involving sulfur:
S(rhombic) + O2(g) → SO2(g) ΔH o rxn = −296.4 kJ/mol
Now, we can use the given reaction for the formation of carbon disulfide (CS2) to find the enthalpy change for this reaction:
CS2 + 3O2(g) → CO2(g) + 2SO2(g) ΔH o rxn = −1073.6 kJ/mol
By rearranging these equations and manipulating the enthalpy changes, we can find the standard enthalpy of formation of carbon disulfide:
CS2 = CO2 - C(graphite) - 2SO2 = -1073.6 kJ/mol - (-393.5 kJ/mol) - 2(-296.4 kJ/mol) = -1073.6 kJ/mol + 393.5 kJ/mol - 2(-296.4 kJ/mol) = -310.3 kJ/mol
Therefore, the standard enthalpy of formation of carbon disulfide (CS2) from its elements is -310.3 kJ/mol.
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The standard enthalpy of formation of carbon disulfide (CS2) is calculated using Hess's Law and given reactions. After rearranging and combining reactions to form CS2 from its elements, the standard enthalpy of formation for CS2 is found to be -87.3 kJ/mol.
Explanation:To calculate the standard enthalpy of formation (ΔHfo) of carbon disulfide (CS2), we use Hess's Law and the given reactions:
C(graphite) + O2(g) → CO2(g), ΔHo = -393.5 kJ/molS(rhombic) + O2(g) → SO2(g), ΔHo = -296.4 kJ/molCS2(l) + 3O2(g) → CO2(g) + 2SO2(g), ΔHo = -1073.6 kJ/molThe standard enthalpy of formation of CS2 is calculated by rearranging the reactions to derive the formation reaction for CS2 from its elements:
Reverse the first reaction and multiply it by 1 so we have C(graphite) forming from CO2(g).Multiply the second reaction by 2 to account for the 2 moles of SO2 forming from S(rhombic) and O2(g).Add these modified reactions to the third given reaction.In doing so, we get:
C(graphite) + 2S(rhombic) + O2(g) → CS2(l), ΔHfo = -1073.6 kJ/mol + (1 x 393.5 kJ/mol) + (2 x 296.4 kJ/mol)
We sum the enthalpy changes of these steps to find the enthalpy of formation for CS2 which is:
ΔHfo(CS2) = -1073.6 kJ/mol + 393.5 kJ/mol + 592.8 kJ/mol = -87.3 kJ/mol
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The volume of a sample of hydrogen gas was decreased from 12.16 L to 5.45 L at constant temperature. If the final pressure exerted by the hydrogen gas sample was 5.09 atm, what pressure did the hydrogen gas exert before its volume was decreased?
Answer : The pressure of the hydrogen gas exert before its volume was decreased will be, 2.28 atm
Explanation :
According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.
[tex]P\propto \frac{1}{V}[/tex]
or,
[tex]PV=k[/tex]
or,
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of the gas = ?
[tex]P_2[/tex] = final pressure of the gas = 5.09 atm
[tex]V_1[/tex] = initial volume of the gas = 12.16 L
[tex]V_2[/tex] = final volume of the gas = 5.45 L
Now put all the given values in this formula, we get the initial pressure of the gas.
[tex]P_1\times 12.16L=5.09atm\times 5.45L[/tex]
[tex]P_1=2.28atm[/tex]
Therefore, the pressure of the hydrogen gas exert before its volume was decreased will be, 2.28 atm
What drives spontaneous reactions?
Select one:
a. decreasing enthalpy and decreasing entropy
b. decreasing enthalpy and increasing entropy
c. increasing enthalpy and decreasing entropy
d. increasing enthalpy and increasing entropy
Answer:
option B
Explanation:
A spontaneous reaction drive favorable when enthalpy is decreasing and the entropy is increasing on the system. If that happens the reaction occurs spontaneously
Answer:
b. decreasing enthalpy and increasing entropy
Explanation:
∆H stands for enthalpy change and ∆S stands for entropy change
Spontaneity depends on the enthalpy and entropy changes of the reaction
∆G = ∆H - T∆S
When ∆H is negative and ∆S is positive
∆G will be negative
For a spontaneous reaction ∆G is negative
If ∆G = 0 then the reaction will be at equilibrium
If ∆G is positive the reaction is non spontaneous.
Decreasing enthalpy (negative) and increasing entropy (positive) will give a negative number for [tex]\Delta G[/tex]
How does an enzyme increase the rate of the chemical reaction it catalyzes?
Answer:
lowers activation energy
Answer:
Explanation:
An enzyme reduces the free energy of activation (EA) of the reaction it catalyzes.
Hydrates that have a low vapor pressure and remove moisture from air are said to be ___. Question 8 options: effloresce hygroscopic solvation anhydrous
Answer:
Hygroscopic
Explanation:
An hygroscopic substance is one that absorbs moisture from the atmosphere and becomes wet. Their ability to remove water from air is less than that of deliquescent substances. Most of the solid hygroscopic substances forms pasty substances and not solutions like the deliquescent compounds.
Examples are sodium trioxonitrate(v), copper(ii) oxide e.t.c
Efflorescence compounds gives off their water of crystallization to the atmosphere.
Hydrates capable of removing moisture from the air due to their low vapor pressure are known as hygroscopic.
Hydrates that have a low vapor pressure and can remove moisture from air are hygroscopic. Substances such as anhydrous calcium chloride and magnesium chloride exhibit hygroscopic properties due to their ability to absorb moisture, ultimately becoming hydrates in the process.
For example, anhydrous calcium chloride mixed with cobalt chloride serves as both a drying agent and an indicator; cobalt chloride is blue when anhydrous and pink when hydrated, thus revealing the condition of the desiccant.
Furthermore, the presence of nonvolatile solutes, such as these hydrates, can lower the vapor pressure of a solution by preventing the evaporation of solvent molecules. The waters of hydration in compounds are loosely bound water molecules that can often be removed through heating, turning hydrates back into their anhydrous form.
Suppose that the pressure of 0.66 L of gas is 424.9 mm Hg when the temperature is 261.2 K. At what temperature is the volume 7.63 L and the pressure 872.15 mm Hg?
Select one:
a. 6200
b. 1.24e4
c. 3010
d. 11.0
Answer:
Option a. 6200 KExplanation:
1) Data:
V₁ = 0.66 literP₁ = 42.9 mmHgT = 261.2 KT₂ = ?V₂ = 7.63 literP₂ = 872.15 mmHg2) Formula:
Combined law of gases:
P₁ V₁ / T₁ = P₂ V₂ / T₂3) Solution:
Solve for T₂:T₂ = P₂ V₂ T₁ / (P₁ V₁)
Substitute:T₂ = 872.15 mmHg × 7.63 liter × 261.2 K / ( 424.9 mmHg × 0.66 liter)
T₂ = 6198 K
Rounding to 2 significant figures, that is 6200 K, which is the first choice.The boiling point of propanoic acid is higher than that of 1-butanol because: propanoic acid has a higher molecular weight than 1-butanol. propanoic acid is more soluble in water than 1-butanol. propanoic acid is a better hydrogen bond donor than 1-butanol. propanoic acid forms hydrogen bonded dimers and 1-butanol does not. 1-butanol forms hydrogen bonded dimers and propanoic acid does not.
Answer:
propanoic acid forms hydrogen bonded dimers and 1-butanol does not.
Explanation:
Propanoic acids will have a higher boiling point because it forms dimers.
The two compounds have hydrogen bonds as their predominant intermolecular bonds. The intermolecular determines a lot about the physical properties of a substance such as it's viscosity, boiling point, melting point etc.
The two compounds have hydrogen bonds which are bonds that occur between between hydrogen and a more electronegative atom. The electronegative atoms are usually oxygen,nitrogen and fluorine.
In a compound of 1-butanol, we have just a single hydrogen bond between the hydrogen on one compound and the oxygen on the hydroxyl group of another one.
For, propanoic acids, dimerization occurs. Here, we have two hydrogen bonds. The alkanoic acid functional group furnishes the bond. This bond forms between the carbonyl group and hydrogen on a compound and the hydroxyl group and another hydrogen on the same compound.
How many moles of sodium chloride are present in 0.50 kg of a 1.0 m solution of sodium chloride in water? mol sodium chloride
Answer:
=0.5 moles
Explanation:
Let us assume that the sodium chloride solution has a density of 1g/cm³.
Therefore the volume of the 0.5 kg of solution will be calculated as follows.
0.5kg into grams=0.5 kg×1000g/kg
=500g
volume= mass/density
=500g/1g/cm³
=500cm³
The solution is 1.0 M which means that 1.0 moles are in 1000 cm³
500cm³ will have:
(500 cm³×1.0 moles)/1000 cm³
=0.5 moles
If the temperature of 9.3 L of a gas at 6.5 ºC falls by 6.1 Cº, what is the new volume of the gas if the pressure is constant?
Select one:
a. 9.10
b. 9.29
c. 0.0133
d. 45.74
Answer:
b. 9.29 L is the new volume of the gas if the pressure is constant.
Explanation:
As per Charles’s law
At constant pressure for a given amount of a gas,
Volume is directly proportional to its temperature.
Thus the expression is [tex]V \propto T[/tex]
[tex]\frac {V}{T} = k[/tex] where k is a constant
When there is a change in the volume and temperature the expression will be
[tex]\frac {V_1}{T_1} = \frac {V_2}{T_2}[/tex]
where [tex]V_1[/tex] and [tex]T_1[/tex] are the initial volume and initial temperature and [tex]V_2[/tex] and [tex]T_2[/tex] are the final volume and temperature.
Plugging in the values given
[tex]\frac {(9.3L)}{(279.5K)}=\frac {V_2}{(279.1K)}[/tex]
[tex]V_2=\frac {(9.3L\times279.1K)}{279.5K} \\\\=9.29L[/tex]
(Answer)
What process occurs when heavier chemicals fall to Earth as dry particles?
acid rain
condensation
deposition
Acid rain is the process that involves heavy chemicals falling to the Earth
as dry particles.
Acid rain is common in areas which have a lot of industries which release
chemicals into the atmosphere. With time, these chemicals accumulate and
falls to earth as dry particles through precipitation.
The acid rain normally contains water which pushes the chemicals down and the water is usually acidic with a pH between 4.2 and 4.4 as a result of the chemicals.
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How many liters of octane are in a shipping container that contains 45 moles?
A. 45 liters
B. 450 moles
C. 2 moles
D. 1,008 liters
A main goal of most environmental scientists is to achieve?
A- biodiversity
B- cost-benefit balance
C- sustainability
D- supply and demand
A main goal of most environmental scientists is to achieve biodiversity.
A solution involves an attraction between the solute, the substance being dissolved, and the solvent, the substance that does the dissolving. This attraction is based on the polarity of the solute and the solvent. If both the solute and the solvent are polar, the solute will dissolve in the solvent. The same can also be said if the solute and the solvent are nonpolar. This results in the formation of a solution. However, if the solute and the solvent have different polarities (i.e., one is polar and the other is nonpolar), they will not dissolve in one another, and a solution will not form. Which substance will dissolve in ethanol
The solute that dissolves in ethanol depends on the polarity of the solute and the solvent. If both the solute and the solvent are polar, then the solute will dissolve in the solvent. Similarly, if both the solute and the solvent are nonpolar, then the solute will also dissolve in the solvent. However, if the solute and the solvent have different polarities, they will not dissolve in one another.
Explanation:A substance can dissolve in a solvent, and form a solution, if the solute and solvent are attracted to each other. For example, water molecules that are held together by hydrogen bonding will dissolve solutes that can also hydrogen bond, like ethanol (CH3CH₂OH). The new hydrogen bonds between the water and the ethanol molecules (solvent-solute attractions) are nearly as strong as the hydrogen bonds in water (solvent-solvent) and ethanol (solute-solute) alone, making the process of solution formation (also called dissolution or dissolving) favorable.
When water mixes with other polar substances, like ethanol, some of the hydrogen bonding between water molecules replace with similar hydrogen bonding with ethanol molecules. Since the electrostatic potential energy is similar, the natural tendency to go towards more dispersion drives the dispersion of ethanol molecules uniformly in water resulting in the solution.
Nonpolar compounds do not dissolve in water. The attractive forces that operate between the particles in a nonpolar compound are weak dispersion forces. However, the nonpolar molecules are more attracted to themselves than they are to the polar water molecules. When a nonpolar liquid such as oil is mixed with water, two separate layers form because the liquids will not dissolve into each other. When another polar liquid such as ethanol is mixed with water, they completely blend and dissolve into one another. Liquids that dissolve in one another in all proportions are said to be miscible. Liquids that do not dissolve in one another are deemed immiscible. The general rule for deciding if one substance is capable of dissolving another is 'like dissolves like'. A nonpolar solid such as iodine will dissolve in nonpolar lighter fluid, but will not dissolve in polar water.
A 1.44 L buffer solution consists of 0.322 M propanoic acid and 0.112 M sodium propanoate. Calculate the pH of the solution following the addition of 0.069 moles of HCl. Assume that any contribution of the HCl to the volume of the solution is negligible. The Ka of propanoic acid is 1.34 × 10-5.
Answer: pH = 4.11
Explanation: pH of the buffer solution is calculated using Handerson equation:
[tex]pH=pKa+log(\frac{base}{acid})[/tex]
pKa is calculated from the given Ka value as:
pKa = - log Ka
[tex]pKa=-log1.34*10^-^5[/tex]
pKa = 4.87
pH of the solution before adding HCl to it:
[tex]pH=4.87+log(\frac{0.112}{0.322})[/tex]
pH = 4.87 - 0.46
pH = 4.41
Now, 0.069 moles of HCl are added to the buffer solution. This added HCl react with base(sodium propanoate) to produce acid(propanoic acid).
Initial moles of acid = 0.322*1.44 = 0.464
initial moles of base = 0.112*1.44 = 0.161
moles of base after reacting with HCl = 0.161 - 0.069 = 0.092
moles of acid after addition of HCl = 0.464 + 0.069 = 0.533
Let's plug in the values in Handerson equation to calculate the pH:
[tex]pH=4.87+log(\frac{0.092}{0.533})[/tex]
pH = 4.87 - 0.76
pH = 4.11
So, the original pH of the buffer solution is 4.41 and after addition of HCl the pH is 4.11 .
The pH of solution following the addition of 0.069 moles of HCl is 4.11.
How to calculate pH?The pH of a buffer solution can be calculated using Handerson equation as follows:
pH = pka + log (base/acid)
pKa of the acid is calculated from the given Ka value as follows:
pKa = - log Ka
pKa = - log 1.34 × 10-⁵
pKa = 4.87
The pH of the solution before adding HCl to it is as follows:
pH = 4.87 + log(0.112/0.322)
pH = 4.87 - 0.46 = 4.41
According to this question, 0.069 moles of HCl are added to the buffer solution.
Initial moles of acid = 0.322 × 1.44 = 0.464Initial moles of base = 0.112 × 1.44 = 0.161moles of base after reacting with HCl = 0.161 - 0.069 = 0.092moles of acid after addition of HCl = 0.464 + 0.069 = 0.533Therefore, the pH of the buffer after adding HCl is:
pH = 4.87 + log(0.092/0.533)
pH = 4.87 - 0.76 = 4.11
Therefore, the pH of solution following the addition of 0.069 moles of HCl is 4.11.
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One mole of hydrogen gas is reacted with oxygen gas to produce water in this reaction:
2 H2 + O2 → 2 H2O
In the lab you actually made 14 grams of water. What is the percent yield?
35.60%
100%
11.11%
77.60%
Answer:
77.6%
Explanation:
Parameters given:
Number of moles of H₂ = 1 mole
Actual yield = 14g
Equation of the reaction:
2H₂ + O₂ → 2H₂O
Solution
From the statement of the problem, we have been given the actual yield that was obtained during the laboratory procedure. Now, we find the theoretical yield and this would lead us to the percentage yield.
Percentage yield = [tex]\frac{actual yield}{Theoretical yield}[/tex] x 100
Theoretical yield:
From the stoichiometeric equation:
2moles of H₂ produced 2 moles of water
Hydrogen gas the limiting reactant and it is in short supply,
Therefore, 1 mole of water would be produced by 1 mole of hydrogen gas.
We use this to estimate the mass of the water that would be produced:
mass of water = number of moles of water x molar mass of water
Molar mass of H₂O = (2x1) + 16 = 18gmol⁻¹
mass of water = 1 mol x 18gmol⁻¹ = 18g
This is the theoretical yield
Percentage yield = [tex]\frac{actual yield}{Theoretical yield}[/tex] x 100
Percentage yield = [tex]\frac{14}{18}[/tex] x 100 = 77.6%
Using the Lewis Dot Structure, what does IF6+ look like?
I need this as soon as possible tytyy
Answer:
This ion is in an octahedral geometry.
See the diagram attached for the Lewis Dot Structure of the iodide hexafluoride cation [tex]\rm {IF_6}^{+}[/tex]. (Created with Google Drawings.)
Note that in this diagram,
A pair of double dots on an atom represent a lone pair of electrons.A single dash represents a single chemical bond.The square bracket and the superscript indicates that this structure is charged.Explanation:
The iodine in [tex]\rm {IF_6}^{+}[/tex] forms an expanded octet. There are twelve valence electrons in total around this atom.
Each of the six fluorine atom needs 8 - 7 = 1 electron to achieve an octet of eight electrons.The iodine atom needs 12 - 7 = 5 electrons to achieve an expanded octet of twelve electrons.The ion carries a positive charge of +1. Atoms in this ion lacks one extra electron.Overall, there needs to be [tex]6 \times 1 + 5 + 1= 12[/tex] more electrons for seven atoms to achieve an octet. They will form half that number of chemical bonds. That's [tex]12 / 2 = 6[/tex] bonds.
Now consider: what will be the geometry of this ion? There are six chemical bonds but no lone pair around the central iodine atom. The six [tex]\mathrm{I-F}[/tex] bonds repel each other equally. They will stay as far apart from each other as possible. As a result, the shape of the ion will be octahedral. Each of the fluorine atoms occupies a vertex.
Acetylcholine causes an end-plate potential by triggering the
Answer:
Acetylcholine causes an end-plate potential by triggering the opening of sodium channels.
Acetylcholine (ACh), a neurotransmitter released by motor neurons, causes an end-plate potential by binding to receptors in the motor end plate and triggering a depolarization process that eventually leads to an action potential.
Explanation:Acetylcholine (ACh) is a neurotransmitter that plays a pivotal role in triggering an end-plate potential within the neuromuscular system. The process starts when an action potential travels down the motor neuron's axon, triggering the release of these neurotransmitters. The ACh then binds to receptors in the motor end plate and initiates a series of events that lead to changes in ion permeability, the influx of sodium ions into the muscles cells, and ultimately a reduction in the voltage difference between the inside and outside of the cell - a process called depolarization. When ACh binds at the motor end plate, this depolarization is called an end-plate potential.
This depolarization then spreads along the muscle fiber membrane, the sarcolemma, creating an action potential, which moves across the entire cell, creating a wave of depolarization. Therefore, the acetylcholine indeed causes the end-plate potential by triggering the release of these neurotransmitters and the subsequent sequence of events.
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A chemical reaction has reached equilibrium when
In 2.00 min, 29.7 mL of He effuse through a small hole. Under the same conditions of pressure and temperature, 9.28 mL of a mixture of CO and CO2 effuse through the hole in the same amount of time. Calculate the percent composition by volume of the mixture. The effusion rate of a gas is proportional to its root-mean-square speed, which is related to its molar mass.
Answer : The percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
And the relation between the rate of effusion and volume is :
[tex]R=\frac{V}{t}[/tex]
or, from the above we conclude that,
[tex](\frac{V_1}{V_2})^2=\frac{M_2}{M_1}[/tex] ..........(1)
where,
[tex]V_1[/tex] = volume of helium gas = 29.7 ml
[tex]V_2[/tex] = volume of mixture = 9.28 ml
[tex]M_1[/tex] = molar mass of helium gas = 4 g/mole
[tex]M_2[/tex] = molar mass of mixture = ?
Now put all the given values in the above formula 1, we get the molar mass of mixture.
[tex](\frac{29.8ml}{9.28ml})^2=\frac{M_2}{4g/mole}[/tex]
[tex]M_2=40.97g/mole[/tex]
The average molar mass of mixture = 40.97 g/mole
Now we have to calculate the percent composition by volume of the mixture.
Let the mole fraction of [tex]CO[/tex] be, 'x' and the mole fraction of [tex]CO_2[/tex] will be, (1 - x).
As we know that,
[tex]\text{Average molar mass of mixture}=\text{Mole fraction of }CO[/tex]
[tex]\text{Average molar mass of mixture}=(\text{Mole fraction of }CO\times \text{Molar mass of } CO)+(\text{Mole fraction of }CO_2\times \text{Molar mass of } CO_2)[/tex]
Now put all the given values in this expression, we get:
[tex]40.94g/mole=((x)\times 28g/mole)+((1-x)\times 44g/mole)[/tex]
[tex]x=0.1894[/tex]
The mole fraction of [tex]CO[/tex] = x = 0.1894
The mole fraction of [tex]CO_2[/tex] = 1 - x = 1 - 0.1894 = 0.8106
The percent composition by volume of mixture of [tex]CO[/tex] = [tex]0.1894\times 100=18.94\%[/tex]
The percent composition by volume of mixture of [tex]CO_2[/tex] = [tex]0.8106\times 100=81.06\%[/tex]
Therefore, the percent composition by volume of mixture of [tex]CO[/tex] and [tex]CO_2[/tex] are, 18.94 % and 81.06 % respectively.
A flexible container at an initial volume of 8.15 L contains 9.51 mol of gas. More gas is then added to the container until it reaches a final volume of 17.9 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Answer : The number of moles of gas added to the container will be, 20.89 mole
Explanation :
According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,
[tex]V\propto n[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 8.15 L
[tex]V_2[/tex] = final volume of gas = 17.9 L
[tex]n_1[/tex] = initial number of moles of gas = 9.51 mole
[tex]n_2[/tex] = final number of moles of gas = ?
Now we put all the given values in this formula, we get the final moles of gas.
[tex]\frac{8.15L}{17.9L}=\frac{9.51mole}{n_2}[/tex]
[tex]n_2=20.89mole[/tex]
Therefore, the number of moles of gas added to the container will be, 20.89 mole