Answer is Carbon dioxide.
The given gases are trace gases. Our atmosphere
has 0.1% of trace gases. Among those trace gases, carbon dioxide level is
highest as 93.49%. Methane has 0.44% and amount of nitrous oxide is 0.07%. But
when considering the whole atmosphere nitrogen gas is the most abundant gas as
78% and next is oxygen as 21%.
Carbon dioxide is the most abundant of the three gases mentioned in the Earth's atmosphere. Nitrogen and oxygen are the most abundant gases overall. Methane is more effective on a per-molecule basis at heating the atmosphere than carbon dioxide.
The question involves identifying which of the three gases carbon dioxide ([tex]CO_2[/tex]), nitrous oxide ([tex]N_2O[/tex]), or methane ([tex]CH_4[/tex]) is most abundant in the Earth's atmosphere. When considering these gases specifically, carbon dioxide is the most abundant. However, it is important to note that the major components of the atmosphere are nitrogen (78.1%) and oxygen (20.9%), with gases such as argon, carbon dioxide, neon, helium, methane, and nitrous oxide present in much smaller amounts.
Regarding the effectiveness at heating the atmosphere, methane is more effective on a per-molecule basis than carbon dioxide due to its ability to absorb more heat. However, carbon dioxide is more abundant and also contributes significantly to the atmospheric warming, playing a crucial role in the Earth's greenhouse effect beside water vapor, the most abundant greenhouse gas.
Write the balanced nuclear equation for alpha decay of polonium−218. include both the mass numbers and the atomic numbers with each nuclear symbol. use the "sup-subscript" button in the answer palette to enter these numbers correctly. greek letters can be accessed in the drop-down menu that says "-select−."
Identify the element that has a ground state electronic configuration of [ar]4s23d104p 3 .
Which phase(s) would experience an increase in the rate of dissolution with an increase in temperature?
A. gases only
B. solids only
C. liquids only
D. both solids and liquids
How many carbon atoms will be in the longest chain 3,3,4,4-tetramethylhexane
What volume of 6.00 m naoh solution is required to prepare 0.50 ml of 0.15 m naoh solution
a. 12.5 ml of naoh
b. 25 ml of naoh
c. 30 ml of naoh?
Final answer:
To prepare 0.50 mL of a 0.15 M NaOH solution from a 6.00 M NaOH solution, you would need 12.5 μL of the concentrated solution according to the dilution formula, which is not listed among the provided answer options.
Explanation:
The volume of 6.00 M NaOH solution required to prepare 0.50 mL of 0.15 M NaOH solution can be found using the dilution formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively. Plugging the known values into the equation: (6.00 M)(V1) = (0.15 M)(0.50 mL) ⇒ V1 = (0.15 M)(0.50 mL) / (6.00 M) = 0.0125 mL or 12.5 μL.
Since none of the provided answer choices (a) 12.5 mL, (b) 25 mL, or (c) 30 mL match this result, it seems there might be a typo or a misprint in the question. To prepare 0.50 mL of a 0.15 M solution from a 6.00 M solution, you would need substantially less than 1 mL of the concentrated solution. The likely correct volume needed, based on the question's data, would be 12.5 μL, which is not listed as an option in the question.
Determine the hydroxide ion concentration in a solution that is 0.0033 m hno3. answer in units of m.
The ___ is the part of the solution that is dissolved and the ___ is the part of the solution that does the dissolving.
What are three ways that carbon atoms can be arranged in hydrocarbon molecules?
Calculate the percent composition by mass of oxygen in litharge (gram formula mss=223.2 grams per mole)
What volume, in milliliters, of 2.0 calcium chloride stock solution would you use to make 500 ml of 0.300 m of calcium chloride cacl2 solution?
Which is the best description of a chain of custody?
A) A list of all personnel entering the crime scene.
B) A log of which family member has ownership of the evidence.
C) A log of which agency controls the evidence so the case stays in the correct jurisdiction.
D) A log of who has handled the evidence and when so the integrity of the evidence is upheld.
how many grams of lithium are needed to produce 45.0 grams of lithium nitride, according to the following process?
6 Li(s) + N2(g) → 2 Li2N(s)
How many grams of solute are present in 615 ml of 0.630 m kbr?
Ka for hcn is 4.9 ⋅ 10-10. what is the ph of a 0.068 m aqueous solution of sodium cyanide? ka for hcn is 4.9 10-10. what is the ph of a 0.068 m aqueous solution of sodium cyanide? 0.74 7.00 2.96 13.24 11.07
Answer
is: pH of solution of sodium cyanide is 11.07.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l)
⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.068 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻]
= 0.068 M - x..
2.04·10⁻⁵ = x² / (0.068 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00116 M.
pOH = -log(0.00116 M) = 2.93.
pH = 14 - 2.93 = 11.07.
The pH of the sodium cyanide solution is 11.56.
Let the cyanide ion be X
We have to set up the ICE table for the problem as follows;
X^-(aq) + H2O(l) ⇄ HX(aq) + OH^-(aq)
I 0.068 x x
C -x + x +x
E 0.068 - x x x
But Kb = Kw/Ka = 1 × 10^-14/4.9 × 10-10
Kb = 2 × 10^-4
So;
Kb = [HX] [OH^-]/[X^-]
2 × 10^-4 = x^2/ 0.068 - x
2 × 10^-4(0.068 - x) = x^2
1.36 × 10^-5 - 2 × 10^-4x = x^2
x^2 + 2 × 10^-4x - 1.36 × 10^-5 = 0
x= 0.0036 M
Since x = [OH^-] = 0.0036 M
pOH = - log(0.0036 M)
pOH = 2.44
pH = 14 - 2.44 = 11.56
The pH of the sodium cyanide solution is 11.56.
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J. J. Thomson
A) Electrons are in orbital levels.
John Dalton
B) Atoms are the smallest unit of an element.
Ernest Rutherford
C) Electrons are scattered within the atom.
Niels Bohr
D) Protons are at the center of an atom.
How is the energy divided up in the plant
Explanation:
Plants veins provide energy to the rest of the plant and it also provides structure and support to the plant leaves and also transport water. When plants absorb water and nutrients through their roots, their vascular system come in use to move water and nutrients to the rest part of the plants. There are two types of tissue that make up the plants veins are xylem and phloem. Xylem moves water and minerals from the plants's roots and phloem moves food energy to all parts of plant where the plants need it.
Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?
Among C₆H₁₂O₆, C₂H₅OH, CH₃COOH, and NaCl, NaCl is the one that will provide the aqueous solution with the lowest freezing point.
We have 4 aqueous solutions and we want to determine which has the lowest freezing point.
C₆H₁₂O₆C₂H₅OHCH₃COOHNaClWhat is the freezing point depression?Freezing-point depression is a drop in the temperature at which a substance freezes, caused when a smaller amount of another, non-volatile substance is added.
We can calculate the freezing point depression (ΔT) using the following expression.
ΔT = Kf × b × i
where,
Kf is the cryoscopic constant for water.b is the molality of the solution.i is the Van't Hoff factor.Assuming all the solutions have the same molality, the freezing point depression will be a function of the van't Hoff factor.
What is the van't Hoff factor?The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
For non-electrolytes, such as C₆H₁₂O₆ and C₂H₅OH, i = 1. Comparing NaCl and CH₃COOH, we can determine that i(NaCl) > i(CH₃COOH) because NaCl is a strong electrolyte and CH₃COOH a weak one.Thus, NaCl, with the highest Van't Hoff factor, will have the lowest freezing point.
Among C₆H₁₂O₆, C₂H₅OH, CH₃COOH, and NaCl, NaCl is the one that will provide the aqueous solution with the lowest freezing point.
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The ph of lemon juice is approximately 2.40. at this ph, the hydronium ion concentration is closest to which concentration? 4.0 × 10¯3 m 0.38 m 5.6 × 10¯4 m 2.5 × 10¯12 m
Help! Brainliest will be given!
What is the product when Bismuth (symbol Bi, atomic number 83, and radioisotope with mass number of 214) emit an alpha particle?
Give the name (or symbol) of the new element, its atomic number, and its mass.
Answer : The correct answer is Name = Thallium (Th) , atomic mass = 210 and atomic number = 81 .
Alpha decay :
When atomic nucleus emits an alpha particle that process is known as alpha particle decay . The symbol of alpha particle is [tex] ^4_2He [/tex] . where 4 is atomic mass ( 2 protons + 2 neutrons ) and 2 atomic number .
When alpha particle is released the resultant nuclei (daughter nuclei) have 4 less atomic mass and 2 less atomic number .
Example for nuclear equation of alpha decay can be expressed as:
[tex] {_{90}^{230}Th} \rightarrow {_2^4He} + {_{88}^{226}Ra} [/tex]
Given :
Radioisotope = Bi Atomic mass = 214 Atomic number = 83
When it will release alpha particle , atomic mass will be decreased by 4 and atomic number by 2 .
Atomic mass = 214 - 4 = 210 Atomic number = 83-2 = 81
The atom with atomic number 81 is Thallium (Th) . Hence the daughter nuclei so produced is [tex] ^2^1^0_8_1Th [/tex]
The nuclear reaction can be written as :
[tex] ^{214}_{83}Bi \rightarrow _2^4He + _{81}^{210}Th [/tex]
Which of the following best describes the changes involved in a fission reaction?
The nucleus of an atom splits into fragments, releasing a large amount of energy.
The nucleus of an atom splits into fragments, absorbing a large amount of energy.
The nuclei of two small atoms combine to form one larger nucleus, releasing a large amount of energy.
The nuclei of two small atoms combine to form one larger nucleus, absorbing a large amount of energy.
In a fission reaction, the nucleus of an atom splits into fragments, releasing a large amount of energy. Nuclear fission occurs when a heavy nucleus like uranium absorbs a neutron and splits, emitting additional neutrons, gamma rays, and energy.
The correct description of the changes involved in a fission reaction is: The nucleus of an atom splits into fragments, releasing a large amount of energy. Nuclear fission is a nuclear reaction where the nucleus of an atom, such as uranium-235 or plutonium-239, splits into smaller nuclei after absorbing a neutron.
This process produces additional free neutrons, gamma radiation, kinetic energy of fission fragments, and releases a significant amount of energy. The process is exothermic, and the energy released can be harnessed for electricity generation in nuclear reactors or can be used in nuclear weapons.
Nuclear fission differs from nuclear fusion, which is the combining of two smaller atomic nuclei to form a larger nucleus, also releasing energy. In both processes, large amounts of heat and radiation are emitted. However, for fission, a fissile material like uranium or plutonium is necessary to sustain the reaction, and it involves the breakup of a heavy nucleus into lighter elements.
Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?
Final answer:
To find the volume of oxygen needed to combust 5.53 grams of propane, calculate the moles of propane, use the stoichiometric relationship to find the moles of oxygen required, and then apply the ideal gas law to find the volume, resulting in 14.79 liters of oxygen needed.
Explanation:
Gas Stoichiometry of Propane Combustion
To determine the volume of oxygen at 25 degrees Celsius and 1.04 atm needed for the complete combustion of 5.53 grams of propane, we need to use stoichiometry and the ideal gas law. The balanced equation for the combustion of propane C3H8 is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
First, we convert the mass of propane into moles using its molar mass:
5.53 g C3H8 × (1 mol C3H8 / 44.11 g C3H8) = 0.1253 mol C3H8
According to the balanced equation, 1 mole of propane reacts with 5 moles of oxygen. Therefore, we need:
0.1253 mol C3H8 × (5 mol O2 / 1 mol C3H8) = 0.6265 mol O2
Now, we use the ideal gas law, PV=nRT, to find the volume of oxygen. First, adjust R to match the pressure in atm and volume in liters:
1.04 atm × V = 0.6265 mol × 0.0821 L×atm/mol×K × 298.15 K
Solving for V gives us the volume of oxygen needed:
V = (0.6265 mol × 0.0821 L×atm/mol×K × 298.15 K) / 1.04 atm = 14.79 L
Therefore, 14.79 liters of oxygen at 25 degrees Celsius and 1.04 atm are required for the complete combustion of 5.53 grams of propane.
What colors of visible light do green plant leave absorb the most
Decreasing the particle size of the reactants in a chemical reaction will increase the rate of reaction by
A) increasing the concentration of the reactants.
B) increasing the surface area of the reactants.
C) increasing the activation energy.
D) increasing the temperature.
Answer: B) increasing the surface area of the reactants.
Explanation:
Decreasing the particle size of the reactants in a chemical reaction will increase the rate of reaction by increasing the surface area of the reactants. Hence, option B is correct.
What is a chemical reaction?A chemical reaction is a process that leads to the transformation of one set of chemical substances into another. It involves breaking and forming chemical bonds between atoms and molecules to form new substances.
In a chemical reaction, the reacting substances are called reactants, and the newly formed substances are called products.
Decreasing the particle size of the reactants in a chemical reaction will increase the rate of reaction by increasing the surface area of the reactants that are available for the reaction.
When the particle size is reduced, the total surface area of the reactant particles is increased, which makes it easier for the particles to collide and react with one another.
Thus, option B is correct.
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What causes a ball rolling across a rug to slow to a stop
Answer:
Friction
Explanation:
Friction is a force that slows down moving objects. If you roll a ball across a shaggy rug, you can see that there are lumps and bumps in the rug that make the ball slow down. The rubbing, or friction, between the ball and the rug is what makes the ball stop rolling.
What is the final pressure of a system (atm) that has the volume increased from 0.75 l to 1.1 l with an initial pressure of 1.25 atm? 1.1 0.85 1.8 1.2 none of the above?
Final answer:
The final pressure of the system is 0.85 atm.
Explanation:
To determine the final pressure of the system, we can use the relationship between volume and pressure known as Boyle's Law. According to Boyle's Law, when the volume of a gas increases, the pressure decreases, and vice versa, at constant temperature and amount of gas.
In this case, the initial volume is 0.75 L and the final volume is 1.1 L. Since the volume increased, we can expect the pressure to decrease. The initial pressure is 1.25 atm. Using Boyle's Law, we can set up the equation:
P1 * V1 = P2 * V2
1.25 atm * 0.75 L = P2 * 1.1 L
Solving for P2, we get: P2 = 1.25 atm * 0.75 L / 1.1 L = 0.85 atm
Therefore, the final pressure of the system is 0.85 atm.
We wish to dilute some 18.4 m h2so4 solution to make 600 ml of a 0.1 m h2so4 solution. how much of the 18.4 m solution should we start with?
We should start with approximately [tex]3.10 ml[/tex] of the [tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution to make [tex]600 ml[/tex] of a [tex]0.1 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution.
To solve this problem, we can use the dilution formula:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
[tex]\( C_1 \)[/tex] is the concentration of the stock solution ([tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex]).
[tex]\( V_1 \)[/tex] is the volume of the stock solution we need to use (unknown).
[tex]\( C_2 \)[/tex] is the concentration of the final solution ([tex]0.1 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex]).
[tex]\( V_2 \)[/tex] is the volume of the final solution ([tex]600 ml[/tex] or [tex]0.6 L[/tex]).
We want to find [tex]\( V_1 \)[/tex], so we rearrange the formula:
[tex]\[ V_1 = \frac{C_2V_2}{C_1} \][/tex]
Now, plug in the values:
[tex]\[ V_1 = \frac{(0.1 \text{ M})(0.6 \text{ L})}{18.4 \text{ M}} \][/tex]
[tex]\[ V_1 = \frac{0.06}{18.4} \][/tex]
[tex]\[ V_1 =0.00326 \text{ L} \][/tex]
Since [tex]1 L = 1000 ml[/tex], we convert [tex]\( V_1 \)[/tex] to milliliters
[tex]\[ V_1 = 0.00326 \text{ L} \times 1000 \text{ ml/L} \][/tex]
[tex]\[ V_1 = 3.26 \text{ ml} \][/tex]
Rounding to two decimal places, we find that we need approximately [tex]3.10 ml[/tex] of the [tex]18.4 M[/tex] [tex]H\(_2\)SO\(_4\)[/tex] solution to make [tex]600 ml[/tex] of a [tex]0.1 M[/tex][tex]H\(_2\)SO\(_4\)[/tex] solution.
When fracking liquid waste is left in pools on the surface, _ can evaporate into the air and contribute to pollution
Volatile organic compounds (VOCs) can evaporate from fracking liquid waste pools, contributing to air pollution. These pools may also contain hazardous air pollutants and heavy metals, requiring careful handling to avoid environmental contamination.
When fracking liquid waste is left in pools on the surface, volatile organic compounds (VOCs) can evaporate into the air and contribute to pollution. These wastewater ponds can contain a variety of pollutants, including hazardous air pollutants such as benzene, toluene, ethylbenzene, and xylene. Moreover, fracking fluid, also known as flowback, can contain chemicals used in the drilling process, heavy metals, and radioactive materials. These substances pose a significant risk to both environmental and human health if they are not properly managed and treated.
The process of hydraulic fracturing or 'fracking' involves injecting high-pressure fluids to fracture shale deposits, which releases trapped gas and oil. The wastewater from this process may return to the water cycle, but the large volume of contaminated water requires careful handling to prevent land and water pollution. As a proactive measure, governments around the world have taken steps, in some cases banning the practice due to the severe risks associated with fracking.
The chemical equation below shows the decomposition of nitrogen triiodide (NI3) into nitrogen (N2) and iodine (I2). 2NI3 mc030-1.jpg N2 + 3I2 The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?
Answer:
0.0135 moles of iodine will be formed.
Explanation:
[tex]2NI_3\rightarrow N_2+3I_2[/tex]
Moles of nitrogen triiodide =[tex]\frac{3.58 g}{394.71 g/mol}=0.0090 mol[/tex]
According to reaction 2 moles of nitrogen triiodide gives 3 moles of iodione gas.
Then 0.0090 mol of nitrogen triiodide will give:
[tex]\frac{3}{2}\times 0.0090 mol=0.0135 mol[/tex]
0.0135 moles of iodine will be formed.
Adding salt to water when boiling vegetables or pasta makes the water boil at a higher temperature. How much would the boiling point of water be increased if 49.2 g of NaCl is added to 500 mL of water? (Remember that the density of water is 1 g/mL and kb for water is 0.51 oC/m)
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
The solubility product constant (Ksp) of calcium carbonate is used to calculate the solubility of this compound in water. Considering the 1:1:1 molar ratio between CaCO3 and the ions it forms on dissolution, we derive the solubility as 0.0058 g/l.
Explanation:The subject of your question is the solubility product constant (Ksp) of calcium carbonate (CaCO3). The Ksp is used to calculate the solubility of a compound, such as CaCO3, in a given solvent, which in this case is water. The Ksp is given as 3.36 × 10-9 M² and it is this value that we will use to determine the solubility.
As we know, for the dissolution of calcium carbonate into calcium and carbonate ions, the dissolution reaction can be written as: CaCO3(s) → Ca²+ + CO3²-. Here, it implies that there is a 1:1:1 molar ratio between CaCO3 and the ions it forms on dissolution. Thus, if 's' represents the molarity of CaCO3, then the Ksp expression can be written as Ksp = [Ca²+][CO3²-] = s².
Using the given Ksp value, we can solve the above equation for 's': s = sqrt(Ksp) = sqrt(3.36 × 10-9 M²) = 5.8 × 10-5 M. However, the question asks for the solubility in g/l. To convert molarity (M) to g/l, we multiply by the molecular weight of the compound. For CaCO3, the molecular weight is approximately 100 g/mol. Therefore, the solubility is 5.8 × 10-5 M x 100 g/mol = 0.0058 g/l.
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