D. methane, butane and propane
Propane and butane are both types of alkane hydrocarbons with relatively low boiling points, which allows them to be easily converted to a liquid form at relatively low pressures. This makes them easy to transport and store. Methane, ethane, and hexane are also alkane hydrocarbons, but they are not typically found in large amounts in LPG.
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True or False: The bowling ball pendulum didn’t hit Bill Nye in the face because the amount of kinetic energy can never be more than the amount of potential energy, some of the energy is transferred to heat.
The bowling ball pendulum didn’t hit Bill Nye in the face because the amount of kinetic energy can never be more than the amount of potential energy, some of the energy is transferred to heat. This statement is true.
Why didn't the bowling ball pendulum hit Bill Nye in the face?The bowling ball pendulum did not hit Bill Nye in the face because potential energy will always be greater than kinetic energy. When a force (a push or a pull) acts on something over a long distance, this occurs.
For instance, if the object is positioned at a higher height, its kinetic energy will be higher. Potential energy is not transferable and varies with object mass, height, and distance.
Thus, The bowling ball pendulum didn’t hit Bill Nye in the face because the amount of kinetic energy can never be more than the amount of potential energy, some of the energy is transferred to heat.
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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA HELP
Answer:
c
c
Explanation:
mark me brainliest pls
Describe how solid ammonium chloride can be separated from a solid mixture of ammonium chloride and anhydrous chloride
Answer: Solid ammonium chloride can be separated from a solid mixture of ammonium chloride and anhydrous chloride through the process of sublimation.
Explanation:
A process in which solid substance converts directly into gas without undergoing liquid state is called sublimation.
For example, when solid ammonium chloride is heat in the apparatus then it will convert directly into vapors without undergoing liquid state. During this process it leaves behind anhydrous chloride into the mixture itself.
When these vapors are cooled down at the surface of funnel then it gives ammonium chloride in the solid state.
Thus, we can conclude that solid ammonium chloride can be separated from a solid mixture of ammonium chloride and anhydrous chloride through the process of sublimation.
How many bonds could each of the following chelating ligands form with a metal ion?
a. acetylacetone(acacH)
b. dirthylenetriamine
c. Salen
d. porphine
a. Acetylacetone(acacH) can form up to 4 bonds with a metal ion.
b. Dirthylenetriamine can form up to 4 bonds with a metal ion.
c. Salen can form up to 6 bonds with a metal ion.
d. Porphine can form up to 8 bonds with a metal ion.
la actual definición de un elemento es
Answer:
La definición de un elemento es una parte o aspecto de algo abstracto, especialmente uno que es esencial o característico.
What is the hydroxide [OH-] concentration of a solution that has a pOH of 4.90? 14 14 1.26 x10-5 1.26 x10, -5 9.1 9.1 7.94 x 104
Answer:
The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.
Explanation:
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution.
POH indicates the concentration of hydroxyl ions [OH-] present in a solution and is defined as the negative logarithm of the activity of hydroxide ions (that is, the concentration of OH- ions):
pOH= -log [OH-]
A solution has a pOH of 4.90. Replacing in the definition of pOH:
4.90= -log [OH-]
Solving:
-4.90= log [OH-]
1.26*10⁻⁵ M= [OH-]
The hydroxide [OH-] concentration of the solution is 1.26*10⁻⁵ M.
The picture below shows an open field with wildflowers.
Which of the following is NOT a way that this environment supports populations of bees?
Answer:
no picture you need add it
1. isostasy subduction 2. deepest ocean depths fjord 3. drowned glaciated valley plate tectonics 4. theory of crustal structure trench 5. oceanic plate going down buoyancy
The isostasy is relates to the buoyancy , the deepest oceans depths relates to the trench, the theory of the crustal structure relates to the plate tectonics.
The match the following for the given options are given below as follows :
1 ) isostasy buoyancy
2) deepest ocean depths trench
3) drowned glaciated valley fjord
4) theory of crustal structure tectonic plates
5) oceanic plate going down subduction
The above are the correct order for the given following relations. these given options are related to the earth movement. like isostasy is the rising or the settling of the part of the earth's lithosphere.
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1.) The property by the virtue of which the metals can be beaten into sheets.
2) Observe the even set up and answer the questions that follow
PLEASE HELP ME
Answer:
1.) The property by the virtue of which the metals can be beaten into sheets.
Malleability
Click it while asking question it allow u to paste image.
At 125 C the pressure of a sample of oxygen (O2) is 2.25 atm. What will the
pressure be at 25°C in closed (constant volume) container?
Answer:
1.68 atm
Explanation:
Applying
P/T = P'/T'................... Equation 1
Where P = Initial pressure, T = Initial Temperature, P' = Final pressure, T' = Final Temperature.
Make P' The subject of the equation
P' = PT'/T.............. Equation 2
From the question,
Given: P = 2.25 atm, T = 125°C = (125+273) K = 398 K, T' = 25°C = (25+273) K = 298 K.
Substitute these values into equation 2
P' = (2.25×298)/398
P' = 1.68 atm
What type of circuit does this figure represent?
parallel circuit
series circuit
short circuit
open circuit
Two lengths of wire connect opposite sides of a battery to contacts on opposite sides of lightbulb one, then lightbulb two, and then lightbulb three.
HELP NOWWWWWWW I WILL MARK BRANILEST!!
Answer:you answer would be B
Explanation:and I'm bc I took the rest lol
Parellel Curcuit Explanation: I took the Test On k12
A flask contains a mixture of N2 and O2 at STP. If the partial pressure of N2 is 40 kPa, what's the mole fraction o
A flask contains a mixture of N₂ and O₂ at STP. If the partial pressure of N₂ is 40 kPa, the partial pressure of O₂ is 61.325 kPa.
STP is the abbreviation which is used for a Standard Temperature and Pressure. The standard temperature is 273 K as well as the standard pressure is 1 atm pressure.
1 atm = 101.325 kPa
According to Dalton's law, state's that the sum of the individual pressures exerted by the gases in a mixture is the total pressure.
[tex]P_{total}[/tex]= [tex]P_{A}[/tex] + [tex]P_{B}[/tex]
Therefore, [tex]P_{total}[/tex]= [tex]p_{n2}[/tex] + [tex]P_{O2}[/tex]
[tex]P_{N2}[/tex] = 40kPa
[tex]P_{total}[/tex] = 101.325kPa
101.325kPa = 40kpa + [tex]P_{O2}[/tex]
[tex]P_{O2}[/tex] = 61.325kpa
--The given question is incorrect, the correct question is
--"A flask contains a mixture of N₂ and O₂ at STP. If the partial pressure exerted by the N₂ is 40.0 kPa, the partial pressure of O₂ is?"--
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If sunlight is breaking down p-nitrophenol molecule, which of the 5 basic
reactions is happening?
Answer:
Chemical reaction, not physical
Explanation:
changing the subscripts of chemicals can mathematically balance the equation. why is it unacceptablle
To balance a chemical equation, you can only adjust the coefficients; you cannot change the subscripts. The ratios of the atoms in the molecule and the resulting chemical characteristics are altered by changing subscripts.
What occurs if the subscript in a chemical equation is changed?Changing the coefficients solely affects the quantity of molecules in that specific chemical. However, modifying the subscripts will change the substance itself, rendering your chemical equation incorrect.
What distinguishes altering a subscript from altering a coefficient in a chemical formula?You can determine the substance from the subscript. You can find out how much of each element is present in the molecule. The substance itself would change if it were altered. In contrast, simply altering the coefficient results in a change in the number of molecules.
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Step 2: measure the area of the top of the syringe
The top of the syringe is a circle. You need to compute area for use in lead to computations of pressure valves. Start by using a ruler to measure the diameter. Estimate to the nearest. 1 cm
3. 60 cm
Divided by two to find the radius. Maintain significant figures
1. 80 cm
Substitute the radius into the formula pie r 2 To find the area of the top of the syringe. Maintain significant figures
10. 2 cm
The syringes' circle measures area 10.17 square inches.
The area of a circle is the area that a circle occupies on a two-dimensional plane. The region contained by the circle's radius or perimeter is another way to define the circle's area. The circle's radius, or r, is equal to its diameter in half.
The diameter given is 3.60cmThe radius given is 1.80cmWe need to calculate the area of the circle
You may determine the circle's surface area via
Change the formula's value for r.
Area of circle =π[tex]r^{2}[/tex]
Area of circle =3.14[tex]\times[/tex]1.80[tex]\times[/tex]1.80
Area of circle =10.17 [tex]cm^{2}[/tex]
Hence the area of the Circle is 10.17[tex]cm^{2}[/tex]
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A gas occupies 43.0 L of space at 225 ℃. Calculate the final temperature of the gas if it is compressed to 20.6 L of space.
Answer:
[tex]T_2=238.57^{\circ} C[/tex]
Explanation:
Given that,
Initial volume, V₁ = 43 L
Initial temperature, T₁ = 225℃ = 498 K
Final volume, V₂ = 20.6 L
We need to find the final temperature. We know that, the relation between volume and temperature is given by :
[tex]V\propto T\\\\\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}[/tex]
Put all the values,
[tex]T_2=\dfrac{498\times 20.6}{43}\\\\T_2=238.57^{\circ} C[/tex]
So, the final temperature is equal to [tex]238.57^{\circ} C[/tex].
Beryllium iodide + strontium sulfate --->
Normal:
Complete ionic:
Net ionic:
Normal
BeI2 + SrSO4 → SrI2 + BeSO4
Complete ionic:
Be2+ + 2I- + Sr2+ + SO42- → Sr2+ + 2I- + Be2+ + SO42-
Net ionic:
Be2+ + 2I- → Sr2+ + 2I- + Be2+
The complete ionic equation for this reaction is the same as the normal equation, except that the reactants and products are written as their respective ions.
The net ionic equation is the same as the complete ionic equation, except that the spectator ions, which are ions that appear on both the reactant and product sides, are omitted.
In this reaction, strontium sulfate and beryllium iodide react to form strontium iodide and beryllium sulfate. The spectator ions are Sr2+ and 2I-, which appear on both sides of the equation and cancel out, leaving the net ionic equation.
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What is the ph and poh of a solution made by adding water to 15 grams of hydrochloric acid until the volume of the solution if 2500 mL
Answer:
pH= 0.8
pOH= 13.2
Explanation:
Number of moles = mass/molar mass
mass= 15 g
molar mass =36.5 g/mol
Number of moles = 15 g/36.5 g/mol = 0.4 moles
From;
n= CV
Where;
C= concentration of solution
V= volume of solution
n= Number of moles
C = n/V
n= 0.4 moles
V= 2500mL or 2.5 L
C= 0.4/2.5
C= 0.16 M
Hence [H^+] = 0.16 M
pH= - log [H^+]
pH= 0.8
pOH= 14 - pH
pOH= 14 - 0.8
pOH= 13.2
can someone label these?
1. Label
A. sediment
B. weathering & erosion
C. sedimentary rock
D. Igneous rock
E. Melting
2. label
A. Sediment
B.weathering & erosion
C. Sedimentary rock
D. Igneous rock
E. Melting
3. Label
A. Sediment
B.weathering & erosion
C. Sedimentary rock
D. Igneous rock
E. Melting
4. Label
A. Weathering & erosion
B. Sedimentary rock
C. Igneous rock
D.Melting
E. Heat &pressure
5. Label
A. Weathering & erosion
B. Sedimentary rock
C. Igneous rock
D.Melting
E. Heat &pressure
6. Label
A. Weathering & erosion
B. Sedimentary rock
C. Igneous rock
D.Melting
E. Heat &pressure
Explain why estuaries, salt marshes, and mangrove forests have more variation in salinity than other marine habitats like the ocean and sea floor.
completely saturated with salt to freshwater. ... Salt marshes, estuaries, and mangrove forests are each unique ecosystems in ... These areas often serve as nursing grounds where young marine life is ... Several reptiles reside in the salt marsh habitat,
What are memory B cells and memory T cell
Answer: Memory B and memory T cells are types of lymphocytes.
Explanation:
Memory B cells forms the part of adaptive immune system. These cells develop inside the germinal centers of the secondary lymphoid organs. These cells circulate in the blood. These are produced by the bone marrow and can be found in tonsil and spleen. Memory B cells are responsible for recognizing previous infections with a foreign antigen and prepare antibody against the new infection. Memory T cells recognize the antigen lived in the body for longer and produce quick response against it on second exposure. These are one of the types of T lymphocytes.
15.0 L of an ideal gas at 298 K and 3.36 bar are heated to 350 K with a new pressure of 4.90 atm. What is the new volume in litres
The new volume in litres is 13.29 L.
Given,
Temperature 1 (T₁) = 298 K
Volume 1 (V₁) = 15 L
Temperature 2 (T₂) = 383 K
Pressure 1 (P₁) = 3.36 atm
Pressure 2 (P₂) = 4. atm
By Ideal Gas Equation,
P₁V₁/T₁ = P₂V₂/T₂
⇒ (3.36*15)/298 = (4.90*V₂)/383
⇒50.4/298 = 4.90V₂/383
⇒ (50.4*383) / (298*4.90) = V₂
⇒ 19303.2/1460.2 = V₂
⇒13.29 L = V₂
New Volume will be 13.29 L.
Ideal gas law is a precise approximation of the behavioral of several gases under various situations in thermodynamics. The Ideal Gas Equation is a mathematical formula that combines empirical laws such as Charle's law, Boyle's law, Gay-law, Lussac's and Avogadro's law.
The equation characterizing the states of hypothetical gases stated mathematically by combinations of empirical or physical constants is known as the Ideal Gas Equation. It's also known as the universal gas equation. It is defined as follows:
"The ideal gas law seems to be the equation of state for a fictitious perfect gas. It is a decent approximation of the behavioral of various gases under numerous situations, while it has significant limitations".
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spectroscopy
There is something strange about Star 3’s spectrum.
Explain what this is called and what it means about this star.
2. How many grams would be in a 2.75 mole sample of carbon (C)?
a. 0.2299
b. 4.379
c. 33.0 g
d. 1.66x10279
Answer:
b is your answer
Explanation:
What are the possible stress or disturbances for chemical reactions
what is the concentration of HCl solution if 20.4 mL of a 1 M NaOH solution was needed to completely react with 25 ml of hcl
The concentration of the HCl solution is 0.816 M (or 816 mM).
To determine the concentration of the HCl solution, we can use the balanced equation for the neutralization reaction between HCl and NaOH:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
This reaction is a neutralization reaction, which means that the number of moles of H+ ions (from the HCl) is equal to the number of moles of OH- ions (from the NaOH).
Given that 20.4 mL of a 1 M NaOH solution was needed to completely react with 25 ml of HCl, we can use the volume and molarity of the NaOH solution to determine the number of moles of OH- ions:
Molarity = moles of solute/litres of solution
Moles of OH- = (20.4 mL x 1 M) / 1000 mL/L = 0.0204 moles
Since the number of moles of H+ ions is equal to the number of moles of OH- ions, we can use the number of moles of OH- to determine the number of moles of H+ ions:
moles of H+ = moles of OH- = 0.0204 moles
We can then use the number of moles of H+ ions and the volume of the HCl solution to determine the concentration of the HCl solution:
Molarity = moles of solute/litres of solution
concentration of HCl = moles of H+ / (25 mL / 1000 mL/L) = 0.0204 moles / 0.025 L = 0.816 M (or 816 mM)
Therefore, the concentration of the HCl solution is 0.816 M (or 816 mM).
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What reagent causes the observed visual change in a positive Lucas test?
The observed visual change from a clear and colourless characteristic to a turbid, cloudy, and hazy characteristic, in a positive Lucas test, is caused by the Lucas reagent.
The Lucas Test is used to detect the presence of a primary alcohol.
When the test is positive, the mixture of the primary alcohol with the Lucas reagent changes from a clear and colourless characteristic to a turbid, cloudy, and hazy characteristic. This visual change is due to the reaction of the primary alcohol with the Lucas reagent which is a mixture of concentrated hydrochloric acid and zinc chloride.
The Lucas Test is used to detect the presence of a primary alcohol, and when a positive result is obtained, the reaction of the primary alcohol with the Lucas reagent, a mixture of hydrochloric acid and zinc chloride, causes the visual change from a clear and colourless characteristic to a turbid, cloudy, and hazy one.
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Coca-Cola is making a version of Coke that uses sucrose (C12H22O11) instead of high fructose corn syrup. If there are 39.24 grams of sucrose in a 12.0 fluid oz. can of this Coke, what is the molarity of sucrose in the Coke
Coca-Cola is making a version of Coke that uses sucrose (C12H22O11) instead of high fructose corn syrup. The molarity of sucrose in the Coke is 0.32 mol/L.
Coca-Cola is making a version of Coke that uses sucrose (C12H22O11) instead of high fructose corn syrup. If there are 39.24 grams of sucrose in a 12.0 fluid oz. can of this Coke, what is the molarity of sucrose in the Coke.
To find the molarity of sucrose in the Coke, we first need to find the number of moles of sucrose present in the can. We can use the mass of sucrose (39.24 g) and the molar mass of sucrose (342.3 g/mol) to find the number of moles of sucrose:
= (39.24 g) / (342.3 g/mol)
= 0.114 mol
Next, we need to find the volume of the can in liters. We can convert the volume in fluid oz. to liters using the conversion factor 1 fluid oz.
= 0.0295735 L:
= (12.0 fluid oz.) x (0.0295735 L/fluid oz.)
= 0.354882 L
Finally, we can use the formula for molarity (moles of solute / liters of solution) to find the molarity of sucrose in the Coke:
= molarity
= (0.114 mol) / (0.354882 L)
= 0.32 mol/L
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A 16. 81 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21. 33 grams of CO2 and 5. 822 grams of H2O are produced. In a separate experiment, the molar mass is found to be 104. 1 g/mol. Determine the empirical formula and the molecular formula of the organic compound
A 16. 81 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21. 33 grams of CO₂ and 5. 822 grams of H₂O are produced. In a separate experiment, the molar mass is found to be 104. 1 g/mol. The empirical formula is CH₃ and molecular formula is C₇H₇O₂₁
The empirical formula can be calculate as follows:
The reaction can be write
CxHyOz → CO₂ + H₂O
As a result, we may state that there are two O-atoms for every one C-atom (12g) in CO2 (32g).
This indicates that the ratio is 12g C/32g O.
The molar mass of CO₂ is 44 g/mole.
(12/44) x 33 = 9 g
There are 9 grams of carbon in every 33 grams of CO₂
The same can be done for water:
In H₂O, each element is represented by two H atoms and one O atom. 2 g H and 16 g O.
[2 g H / 18 g H2O] are present in 5. 822 g of water
(2/18)x 5.822= 0.6468 g
The initial sample had 14.16 g of sample, 9 g of C, and 0.6468 g of H, which resulted in 34.3532g of oxygen.
Determine the amount of moles.
C: 9.0 g / 12.0 g/mol = 0.75 mol
H: 0.6468 g / 1.0 g/mol = 0.6468 mol
O: 34.3532g / 16.0 g/mol = 2.14 mol
Then we divide by the smallest number
C: 0.75 mol / 0.6468 mol = 1
H: 0.6468 mol / 0.6468 mol= 1
O: 2.14 mol/ 0.6468 mol = 3
so the empirical formula will be CH₃
Determine the molecular formula
mass molecular formula / mass of empirical formula = n
We have to multiply the empirical formula by n to get the molecular formula.
104.1 /15 = 7= n
This means the molecular formula = C₇H₇O₂₁
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De acuerdo con la Ley de la Conservación de la Materia postulada por Lavoisier, ¿cuál ecuación química está balanceada? A C2H5OH(l) + O2(g) → CO2 + H2O B C2H5OH(l) + O2(g) → 2CO2 + H2O C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O D C2H5OH(l) + O2(g) → 2CO2 + 3H2O E 2C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
Answer:
C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
Explanation:
La ley de conservación de la materia implica que en una reacción química los átomos en los reactivos son iguales a la cantidad de átomos en los productos:
A C2H5OH(l) + O2(g) → CO2 + H2O
En los reactivos hay dos átomos de carbono pero en los productos solo 1. De esta manera, no cumple la ley de conservación.
B C2H5OH(l) + O2(g) → 2CO2 + H2O
En los reactivos hay 6 átomos de hidrógeno pero en los productos solo 2. No cumple la ley de conservación.
C C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
En los reactivos y productos hay: 2 átomos de carbono, 6 átomos de hidrógeno y 7 átomos de oxígeno. Cumple la ley de conservación.
D C2H5OH(l) + O2(g) → 2CO2 + 3H2O
En los reactivos hay 3 átomos de oxígeno y en los productos 7. No cumple la ley de conservación.
E 2C2H5OH(l) + 3O2(g) → 2CO2 + 3H2O
En los reactivos hay 4 átomos de carbono y en los productos solo dos. No cumple la ley de conservación.