In the process of heating a solid and observing a phase change, the parts of the curve can be sorted based on whether the average energy of the molecules is changing or is constant.
The parts of the curve where the average energy of the molecules is changing include:The solid is heated at the melting point: In this phase, the solid absorbs heat energy, causing the average energy of the molecules to increase as the temperature rises. The solid undergoes a phase change from a solid to a liquid.
The liquid is heated to reach the boiling point: During this phase, the liquid continues to absorb heat energy, leading to an increase in the average energy of the molecules as the temperature rises. The liquid approaches its boiling point, preparing for the phase change into a gas.The parts of the curve where the average energy of the molecules is constant include:The solid is at the melting point: At this stage, the solid remains at a constant temperature as it undergoes the phase change from a solid to a liquid. Although heat is still being added, the extra energy is being used to break the intermolecular forces and convert the solid into a liquid.
The liquid is at the boiling point: Here, the liquid also maintains a constant temperature as it undergoes the phase change from a liquid to a gas. The heat energy supplied is being utilized to break the intermolecular forces and convert the liquid into a gas.By observing the changes in temperature and the corresponding phase changes, we can determine whether the average energy of the molecules is changing or is constant throughout the heating process.
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The reaction between A2 (two large red spheres) and B2 (two small blue spheres) to produce AB is shown in the diagram. The reactants contain 4 A 2 molecules and 3 B 2 molecules. The products contain 6 A B molecules. Determine the limiting reactant for the reaction. AB A2 B2 no limting reactant
Answer:
B₂
Explanation:
The limiting reactant is always a reactant. You can determine which reactant is limiting by identifying which has the smaller mole-to-mole ratio with the product. This ratio can be found via the coefficients of the balanced reaction.
4 A₂ + 3 B₂ ---> 6 AB
4 moles A₂
------------------ = mole-to-mole ratio A₂/AB
6 moles AB
3 moles B₂
------------------ = mole-to-mole ratio B₂/AB
6 moles AB
Since the mole-to-mole ratio between B₂ and AB is smaller, B₂ must be the limiting reactant.
Phosphate buffer solution is purchased as a(n) 10x stock solution. The solution is more concentrated than the working stock. If tasked with making 941 mL of a 1x solution of PBS, how many mL of the stock solution would be needed
94.1 mL of 10x PBS should be needed
Given:
Phosphate buffer solution = 10x
Volume of 1x solution of PBS = 941 mL
To Find:
Volume of stock solution
Solution:
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount of strong acid or base is added to it.
using the formula
M1 x V1 = M2 x V2
(10x)* V1 = 1x* 941
V1 = 94.1 ml
So, 94.1 mL of 10x PBS should be needed
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Which statement is true about real gases with the correct reason:
a.
Molecules have volume which causes deviations from ideal gases at high temperature .
b.
Molecules have volume which causes deviations from ideal gases at low temperature.
c.
Molecules have volume which causes deviations from ideal gases at low pressure .
d.
Molecules have volume which causes deviations from ideal gases at high pressure.
Volume of a real gas shows deviation from ideal gas on low temperature and high pressure
Only 2 options are correct then
Molecules have volume which causes deviations from ideal gases at low temperature.Molecules have volume which causes deviations from ideal gases at high pressure.There are 2 possible answers here : b and d.
The Ideal Gas Equation is : PV = nRT
Here, when pressure is increased and temperature is lowered, the volume of the molecules will substantially decrease, which means it has deviated from ideal behavior.
HELPPP!!! how do i solve this
8. What mass of sodium hypochlorite (NaClO) is required to prepare 1.5 L of chlorine bleach, if the bleach is a 0.70 mol/L solution of sodium hypochlorite? Show your work and explain each step. The molar mass of sodium hypochlorite is 74.44 g/mol.
9. What is the final concentration when 420 mL of a 12.0 mol/L sodium hydroxide solution is diluted to 2.0 L? Show your work and explain each step.
Taking into account the definition of molarity, molar mass and dilution:
the mass of sodium hypochlorite (NaClO) required is 78.162 grams.the final concentration of a sodium hydroxide solution is 2.52 mol/L.Definition of molarityMolar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:
[tex]molarity=\frac{number of moles}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{L}[/tex].
Definition of molar massThe molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
Mass of sodium hypochlorite (NaClO)If the bleach is a 0.70 mol/L solution of sodium hypochlorite and is required to prepare 1.5 L of chlorine bleach, replacing in the definition of molarity you get:
0.70 [tex]\frac{moles}{L}[/tex]=[tex]\frac{number of moles}{1.5 L}[/tex]
Solving:
0.70 [tex]\frac{moles}{L}[/tex]× 1.5 L= number of moles
1.05 moles= number of moles
The molar mass of sodium hypochlorite is 74.44 g/mol. So you can apply the following rule of three: If by definition of molar mass 1 mole of the compound contains 74.44 grams, 1.05 moles of the compound contains how much mass?
[tex]mass=\frac{1..05 molesx74.44 grams}{1 mole}[/tex]
mass= 78.162 grams
Finally, the mass of sodium hypochlorite (NaClO) required is 78.162 grams.
DilutionDilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.
In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.
A dilution is mathematically expressed as:
Ci×Vi = Cf×Vf
where
Ci: initial concentrationVi: initial volumeCf: final concentrationVf: final volumeFinal concentration of sodium hydroxide solutionIn this case, you know:
Ci= 12 mol/LVi= 420 mLCf= ?Vf= 2 L= 2000 mL (being 1 L= 1000 mL)Replacing in the definition of dilution:
12 mol/L× 420mL= Cf× 2000 mL
Solving:
(12 mol/L× 420mL)÷ 2000 mL= Cf
2.52 mol/L= Cf
In summary, the final concentration of a sodium hydroxide solution is 2.52 mol/L.
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A balloon filled with air has a volume of 4.24 liters at 23.00°c. if the balloon is cooled at constant pressure to 5.00°c, what is its new volume? the balloon’s volume at 5°c is liters.
The new volume if the balloon is cooled at constant pressure is 3.98 L.
Charles's law, states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature if the pressure remains constant.
The new volume is calculated using the Charles law formula
V₁ / T₁ = V₂ / T₂
where,
V₁ = The initial volume of air = 4.24 l
T₁ = 23.00 °C into kelvin = 23 +273 =296 K
T₂ = 5.00 °C into kelvin = 5.00 + 273 = 278 K
V₂ = ?
By making V₂ subject the subject of the formula by multiplying both sides by T₂
V₂ = ( V₁ × T₂ ) / T₁
V₂ = (4.24 L × 278 K) / 296 k
= 3.98 L
Therefore, the new volume, if the balloon is cooled at constant pressure, is 3.98 L.
An air-filled balloon will contract when chilled and expand when heated. This occurs because the gas that makes up the air within the balloon expands when it is warm and contracts when it is cool.
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Answer: at 5 its ... 3.98
Explanation:
below shows the dimensions of two colored cubes.
Dimensions of Cubes
Cube Side (cm) Mass (g)
Red: 4 12
Green: 3 10
Which cube is denser?
Red, because it has more volume and more amount of matter
Green, because it has less volume and less amount of matter.
Red, because it has less volume and more amount of matter
Green, because it has less volume and more amount of matter.
Green, because it has less volume and less amount of matter.
What is density?Density is the mass of a material substance per unit volume. d = M/V, where d is density, M is mass, and V is volume, is the formula for density. Grams per cubic centimeter are a typical unit of measurement for density.
Because it enables us to predict which compounds will float and which will sink in a liquid, density is a crucial notion. If an object's density is lower than the liquid's, it will often float.
The dimensions of two colored cubes.
Red: 4 12
Green: 3 10
density = mass/volume
so compare
[tex]12/4^{3}[/tex] and [tex]10/3^{3}[/tex]
None of the choices is the real reason. But is the closest. It is green because 10/27 > 12/64
But red would have been denser for a side of 3.9, and green would still have had less volume and less matter.
Green, because it has less volume and less amount of matter.
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The name of Ra3N₂ is
Answer:
Radium Nitride
What is radium?
A radioactive element is radium. See what "radioactive" actually means.
Atoms make up the elements. Since most atoms are stable, they do not change over time. However, some of the heaviest atoms disintegrate and transform into different types. "Radioactivity" is the term for this breakdown or disintegration.
Each radioactive element releases rays that decay or disintegrate at a specific pace. There is no known way to speed up or slow this rate. Some changes happen quickly, some take longer, but no matter what, man cannot control the action.
In the case of radium, this decay would continue indefinitely until the radioactive material was ultimately converted to lead. For instance, in 1,590 years, half a gram of radium would transform into atoms with a lower atomic weight. Half of the remaining radium would transform after an additional 1,590 years, and so on until all of it turned into lead.
Attachment #1:
This is a revigator. It is a pot that was created specifically with radium, uranium, and other radioactive materials to contain water. When radioactivity was first discovered in 1912, people thought it was beneficial and "magical," similar to how homeopathic treatments are now. They would store water in it to allow it to absorb radiation, then they would drink the water. In the 1930s, several gadgets were subsequently recalled.
The enigmatic new radioactive elements captivated people's curiosity during the early 1900s. Although this starch didn't actually contain radium, it was given that name because of the rarity and beauty of radium. It was once thought of similarly to how titanium is now.
Madame Curie and her husband, Pierre Curie, made the discovery of radium. They were processing a ton of pitchblende, an ore that contains uranium. Although they were aware that the uranium was emitting invisible rays, they had the impression that there was another another, far more potent chemical present. They initially discovered polonium, another radioactive element, before successfully isolating a minute particle of radium.
The three types of rays that radium emits are alpha, beta, and gamma rays. Helium gas contains swiftly moving particles called alpha rays. In beta rays, the electrons move quickly. Gamma rays are similar to X-rays but typically more invasive. Every time one of these rays is emitted, its parent atom transforms into a different element. "Atomic transmutation" is the term for this transformation.
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>> ROR
If 100 pounds of chlorine gas is required for an operation, how many pounds of calcium hypochlorite (65%) would be needed for the same operation
The amount, in pounds of chlorine that will be needed, will be 201.37 pounds
Percent compositionCalcium hypochlorite goes by the chemical formula, [tex]Ca(ClO)_2[/tex].
In order to know the amount of calcium hypochlorite that will be needed for the operation, we first need to know the percent composition of chlorine in calcium hypochlorite.
Molar mass of calcium hypochlorite = 142.98 g/mol
There are 2 molecules of Cl in calcium hypochlorite. The molar mass will be 35.5 x 2 = 71 g/mol
Percent chlorine in calcium hypochlorite = 71/142.98 x 100 = 49.66%
So, what amount of calcium hypochlorite will give us 100 pounds of chlorine knowing that chlorine is 49.66% of calcium hypochlorite?
49.66% of X = 100 pounds
X = 100 x 100/49.66 = 201.37 pounds
Thus, 201.37 pounds of calcium hypochlorite will give us approximately 100 pounds of chlorine.
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Lorelai discovers a compound that is 64.8 g C, 13.62 g H, and 21.58 g O. What is the empirical formula of Lorelai's compound
The empirical formula of Lorelai's compound is C₄H₁₀O
Data obtained from the questionC = 64.8 gH = 13.62 gO = 21.58 gEmpirical formula =? How to determine the empirical formulaDivide by their molar mass
C = 64.8 / 12 = 5.4
H = 13.62 / 1 = 13.62
O = 21.58 / 16 = 1.35
Divide by the smallest
C = 5.4 / 1.35 = 4
H = 13.62 / 1.35 = 10
O = 1.35 / 1.35 = 1
Thus, the empirical formula of the compound is C₄H₁₀O
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What kind of energy is this?
nuclear
radiant
thermal
chemical
The energy can be identified by their features, characteristics and impacts on the surrounding environment.
How to identified kind of energy?If the energy release heat then we can say that it is a thermal energy because it warms the environment whereas if the energy produce light then it is called radiant energy.
If the energy is released from splitting atoms or fusion of atoms so we can say that it is a nuclear energy. If the energy is released from breaking of bonds of atoms and molecules, so it is Chemical energy.
So we can conclude that the energy can be identified by their features, characteristics and impacts on the surrounding environment.
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Tin and oxygen can combine to form two compounds of different combining proportions. Resulting formulas of these compounds include
a. Sn2O and SnO2 d. SnO and SnO3
b. SnO and SnO2 e. SnO2 and SnO4
c. SnO4 and SnO
???
Answer:
B.) SnO and SnO₂
Explanation:
Tin can either exist as Sn²⁺ or Sn⁴⁺ when ionized. Oxygen always exists as O²⁻ when ionized.
When ions combine, sometimes there needs to be different amounts of each ion in order to create an overall charge of 0. In one combination of the ions (Sn²⁺ and O²⁻), there only needs to be one atom of each ion. In another combination (Sn⁴⁺ and O²⁻), you need to have more anions to balance the high positive charge of the cation.
SnO
-----> Sn²⁺ and O²⁻
-----> +2 + (-2) = 0
SnO₂
-----> Sn⁴⁺ and O²⁻
-----> +4 + (-2) + (-2) = 0
4. The equation for the decomposition of hydrogen peroxide is shown. 2H₂O₂(aq) → 2H₂O(l) + O₂(g) 48.0 25.0 cm³ of aqueous hydrogen peroxide forms 48.0 cm³ of oxygen at room temperature and pressure (r.t.p.). Calculate the concentration i aqueous hydrogen peroxide at the start of the experiment using the following steps ● Calculate the number of moles of oxygen formed. • Deduce the number of moles of hydrogen peroxide that decomposed. • Calculate the concentration of hydrogen peroxide in mol/dm³.
Answer:
Just add what they equal to hope this helps :)
Explanation:
Pls Help!
A sample of seawater from a tidal estuary was found to contain a concentration of 727 mg of chloride ion per kg of seawater. If the density of the sample was 1.035 g/mL, what is the molarity of the chloride ion?
The molarity of the chloride ion is 2.0 * 10⁻⁵ M.
What is the molarity of the chloride ion?The molarity of the chloride ion is calculated using the formula below:
Molarity = (Percentage concentration * Density ) / (Molar mass * 100)The percentage concentration of chloride ion = 727/kg * 10⁻⁶ Kg * 100/1
The percentage concentration of chloride ion = 0.0727%
Molar mass of Chloride ion = 35.5 g/mol
Molarity = (0.0727 * 1.035)/(35.5 * 100)
Molarity of chloride ion = 2.0 * 10⁻⁵ M
In conclusion, the molarity of the chloride ion is obtained from the density and percent concentration of chloride ions in seawater.
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A scientist observes light from a star with a wavelength 1 of 97 nanometers (9.7 x 108 m). What is the photon energy, E, of this light?
Use this information to help you:
v=£
speed of light in a vacuum, c = 3.00 x 108 meters/second
• Planck's constant, h = 4.136 × 10-15 ev-s
energy of the photon =
Answer:
1.28 x 10⁻¹⁵ eV
Explanation:
To find the energy, you can use the photon energy equation:
E = hc / λ
In this equation,
-----> E = energy (eV)
-----> h = Plank's Constant (4.136 x 10⁻¹⁵ eV*s)
-----> c = speed of light (3.00 x 10⁸ m/s)
-----> λ = wavelength (m)
You can find the energy of the photon by plugging the given values into the equation and solving.
E = hc / λ
E = (4.136 x 10⁻¹⁵ eV*s)(3.00 x 10⁸ m/s) / 9.7 x 10⁸ m
E = 1.2408 x 10⁻⁶ / 9.7 x 10⁸ m
E = 1.28 x 10⁻¹⁵ eV
Managing large amounts of data and processing them quickly during communication between Earth and space is challenging. Which technology is most suitable to overcome this challenge
Utilizing wireless communication and camera imagery is the most effective technological solution to this problem.
To find the answer, we need to know more about the wireless communication and camera imagery
Wireless communication is what?The transmission of data or information between two or more sites without the use of an electrical conductor is known as wireless communication.It is difficult to manage enormous amounts of data and analyze them fast when communicating between Earth and space.Today, wireless connections are used to access the internet.Thus, we can conclude that, the wireless connectivity and camera imaging method meet this obstacle.
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How many moles of ammonia are produced when 5.0 moles of hydrogen react with excess nitrogen?
Taking into account the reaction stoichiometry, 3.33 moles of ammonia are produced when 5.0 moles of hydrogen react with excess nitrogen.
Reaction stoichiometryIn first place, the balanced reaction is:
N₂ + 3 H₂ → 2 NH₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
N₂: 2 moles H₂: 3 molesNH₃: 2 molesMass of ammonia formedThe following rule of three can be applied: if by reaction stoichiometry 3 moles of hydrogen form 2 moles of ammonia, 5 moles of hydrogen form how many moles of ammonia?
[tex]moles of ammonia=\frac{5 moles of hydrogenx 2 moles of ammonia}{3 moles of hydrogen}[/tex]
moles of ammonia= 3.33 moles
Finally, 3.33 moles of ammonia are produced when 5.0 moles of hydrogen react with excess nitrogen.
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According to the Arrhenius definitions of acids and bases, choose the bases from the list of acids and bases. Check all that apply.
Answer:
a
d
please give me brainliest
Answer:
A C D. All the other answers got it wrong for Edg3
Explanation:
HELP PLS PLS PLS ILL MARK U BRAINLIEST
Answer:
2 FeCl3 → 2 Fe + 3 Cl2
Explanation:
2 Fe and 6 Cl on the reactants side, and 2 Fe and 6 Cl on the products side.
Suppose you need 0.12L of a 0.13M solution of NiCl2, but all you have is a solution that is a solution that is 0.23M. What volume of the solution do you need to dilute
- 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
The dilution problem uses the equation :
[tex]M_aV_a= M_bV_b[/tex]
The initial molarity (concentration) [tex]M_a =[/tex] 0.13 M
The initial volume [tex]V_a[/tex] = 0.12 L
The desired molarity (concentration) [tex]M_b[/tex] = 0.23 M
The volume of the desired solution [tex]V_b[/tex] = ( 0.12 + x L )
Substituting values in above equation;
(0.13 M ) (0.12 L) = (0.23 M ) (0.12 L + x L)
0.0156 M L = 0.0276 M L + 0.23 x M L
- 0.012 M L = 0.23 x M L
x = - 0.05
Therefore, - 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
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Potassium sulfate is added to water. The equation that best represents this process
is:
K₂S(s) → K(s) + 2 S(s)
K₂S(s) K(s) + 2 S(s)
K₂SO4(s) → 2 K (aq) + SO4² (aq)
K₂S(s) 2 K (aq) + S² (aq)
K₂SO4(s) 2 K(s) + SO4(g)
K₂S(s)→2 K (aq) + S² (aq)
K₂SO4(s)→2 K(s) - SO4(g)
K₂SO4(s) 2 K (aq) + SO4²(aq)
The equation that best represents the addition of potassium sulfate to water is [tex]K_2SO_4(s) -- > 2 K^+ (aq) + SO4^{2-} (aq)[/tex]
Dissolution of potassium sulfate in waterWhen potassium sulfate is dissolved in water, it dissociates into its cation and anion according to the following equation:
[tex]K_2SO_4(s) -- > 2 K^+ (aq) + SO4^{2-} (aq)[/tex]
The positive ions are 2 moles of potassium ions while the negative ion is a mole of sulfate ion.
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In an aqueous solution containing Mn(II) and Mn (IV) salts, which cation would you expect to be the more strongly hydrated
In an aqueous solution containing Mn(II) and Mn (IV) salts, Mn (IV) cation is expected to be the more strongly hydrated.
A chemical event called hydration occurs when two substances interact with water. Since Mn (IV) has a stronger cation with a +4 charge than Mn (II), it will draw more oxygen ions from the Mn(II) in aqueous solution. Mn(IV) will consequently be more intensely hydrated.
Another reason for this is that Mn(IV) ions are often smaller than ions. Now, if we take a look at the element that significantly influences hydration, it is ion size. Therefore, the ion will be more hydrated if it is smaller.
So, In an aqueous solution containing Mn(II) and Mn (IV) salts, Mn (IV) cation is expected to be the more strongly hydrated.
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How many moles of H2O are in 64.0 g of H2O
Answer:
Explanation:
stoichiometry of C₂H₂ to H₂O is 2:2.
Number of moles of C₂H₂ = molar mass of C₂H₂
Since the molar mass of C₂H₂ is 26 g/mol.
Number of C₂H₂ moles reacted = 64.0 g / 26 g/mol = 2.46 mol.
according to a molar ratio of 2:2.
the number of H₂O moles formed = a number of C₂H₂ moles reacted.
Therefore the number of H₂O moles produced = 2.46 mol
Which of the following choices represents a balanced chemical equation?
Answer:
the correct answer: 2Ca3(PO4)2 + 6SiO2 -> P4O10 + 6CaSiO3
Explanation:
What is the temperature change the H2O sample
Answer:
5.0 °C
Explanation:
To find the temperature change, you need to use the following equation:
Q = mcΔT
In this equation,
-----> Q = heat energy (J)
-----> m = mass (g)
-----> c = specific heat capacity (J/g°C)
-----> ΔT = temperature change (°C)
You can find the temperature change by plugging the given values into the equation and simplifying.
Q = 209 J c = 4.18 J/g°C
m = 10.0 g ΔT = ? °C
Q = mcΔT <----- Equation
209 J = (10.0 g)(4.18 J/g°C)ΔT <----- Insert values
209 J = (41.8)ΔT <----- Multiply 10.0 and 4.18
5.0 = ΔT <----- Divide both sides by 41.8
Answer:
5° C temp change
Explanation:
Q = m c T
Given: Q =209 m = 10 c = 4.18 looking for T (change in temp)
209 = 10 * 4.18 T
T = 209 / (10*4.18) = 5° C
Which way does heat flow?
Ammonium hydrogen sulfide (NH4HS) can decompose to ammonia and hydrogen sulfide: What is the equilibrium constant expression for this system
The equilibrium constant expression for this system is Keq=[NH3][H2S]
What is the the equilibrium constant expression of the system?The ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation is K for a system in equilibrium, according to the law of mass action. The equilibrium constant expression is known as the ratio.
Therefore, the equilibrium constant expression for this system is Keq=[NH3][H2S]
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Use Table B in your Student Guide to answer the questions about ion concentrations.
A solution with a pH = 13 has approximately how many moles of OH– ions per liter?
0.1
How many moles of H+ would this same solution have per liter?
0.000000000000.1
⇒ 0.0000000000001
(Use the decimal form of your answer.)
A different solution with an H+ concentration of 1.0 × 10–4 would have a pH =
4
.
1. The amount of OH¯ present in the solution is 0.1 mol/L
2. The amount of H⁺ present in the solution is 0.0000000000001 mol/L
3. The pH of the solution with H⁺ concentration of 1×10⁻⁴ mol/L is 4
1. How to determine the amount of OH¯ in the solutionWe'll begin by obtaining the pOH of the solution
pH = 13pOH =?pH + pOH = 14
13 + pOH = 14
Collect like terms
pOH = 14 - 13
pOH = 1
Finally, we can determine the OH¯
pOH = 1OH¯ =?pOH = –Log [OH¯]
1 = –Log [OH¯]
Multiply through by –1
–1 = Log [OH¯]
Take the anti-log of –1
[OH¯] = anti-log of (–1)
[OH¯] = 0.1 mol/L
2. How to determine the amount of H⁺pH = 13H⁺ =?pH = –Log [H⁺]
13 = –Log [OH¯]
Multiply through by –1
–13 = Log [H⁺]
Take the anti-log of –13
[H⁺] = anti-log of (–13)
[H⁺] = 0.0000000000001 mol/L
3. How to determine the pHH⁺ = 1×10⁻⁴ mol/LpH =?pH = –Log [H⁺]
pH = –Log 1×10⁻⁴
pH = 4
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In the reclassify transactions screen, which 3 elements of a transaction can be reclassified?
The 3 elements of a transaction that can be reclassified are:
Account, Location ClassOn an account transaction screen, one can see there are different elements that exist and some of them are:
invoices, sales receipts, checks, or bills, etcWhat is a Transaction?This refers to the proof that exists that a person has bought a good or service and can also stand as an agreement.
Hence, we can see that On an account transaction screen, one can see there are different elements that exist and some of them are:
invoices, sales receipts, checks, or bills, etcHowever, there are only 3 elements that can be reclassified on the reclassify transactions screen and they are:
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1. A gas occupies 0.60 m3 at a 5.0 atm. If the temperature of the gas remains the same and the pressure decreases to 2.5 atm, what would be the new volume occupied by the gas
The new volume occupied by the gas is 1.2 m³.
Here Boyle's law is applied.
What is Boyle's law?At a constant temperature, the volume of a gas is inversely proportional to the pressure exerted on it.
PV = constant
where,
P = Pressure applied
V = Volume of the gas
Calculation:Given,
P₁ = 5.0 atm
V₁ = 0.60 m³
P₂ = 2.5 atm
To find
V₂ =?
As PV = constant
Therefore,
P₁ V₁ = P₂ V₂
5 × 0.6 = 2.5 × V₂
3 = 2.5 × V₂
V₂ = 3/2.5
V₂ = 1.2 m³
Hence, the new volume occupied by the gas is 1.2 m³.
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Increasing the catalytic stability by optimizing the formation of zeolite-supported Mo carbide species ex situ for methane dehydroaromatization
Answer:
The synthesis of zeolite-supported Mo carbide species was studied by testing different reduction/carburization conditions applied to a zeolite-supported Mo oxide catalyst, with the aim to find the optimized treatment conditions necessary to form stable supported Mo carbide catalysts ex situ for application in methane dehydroaromatization reaction. Four types of treatment were performed and studied using temperature-programmed reduction and carburization profiles: (1) heating the catalyst in a reducing gas, H2, up to reaction temperature and switching to CH4; (2) heating the catalyst in a reducing gas, H2, mixed with dilute CH4; (3) heating the catalyst in CH4 up to reaction temperature; and (4) heating the catalyst in an inert gas (commonly He) up to reaction temperature and then switching to CH4 or to H2 followed by CH4 or to H2/CH4 mixture. Each of these processes were stopped at intermediate points to analyze the phases that were present in order to identify the structural evolution of the supported Mo carbides that originate from the supported Mo oxides. Once the supported carbides were formed, they were quenched under the same gas mixture, and then they were each tested in methane dehydroaromatization via previous heating to reaction temperature in He flow. Despite all of them showing only presence of Mo2C species on HZSM-5, the catalytic properties were dramatically different. Catalysts treated in H2 or CH4/H2 showed remarkably higher stability. These catalysts exhibited a higher Mo dispersion and thus exposure on the active surface.
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