Explanation:
The number of electron density regions that encircle an atom is calculated by the steric number, which also defines the hybridisation of the central atom. In that scenario, there are six hybridised orbitals in the central atom, leading to a [tex]\displaystyle sp3d^2[/tex]hybridization.
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The mixing of several atomic orbitals to form the same number of equivalent hybrid orbitals
Answer:
Hybridization
Explanation: -
Hybridization occurs when atomic orbitals mix to form a new atomic orbital
While chemical buffer systems like the bicarbonate-carbonic acid buffer system are primary lines of defense in preventing pH imbalances, what are other organs are associated with this defense?
A. Lung and Kidneys
B. Lungs and Stomach
C. Lungs and Liver
D. Lungs and Intestinal tract
Answer:
Lung and Kidneys are associated with this defence.
Explanation:
An aqueous solution that resists change because of the addition of an acid or base solution is known as a buffer. Thus, it comprises of conjugate base and a weak acid. The bicarbonate buffer system is the one that is associated with the respiratory system.
To maintain the pH of and duodenum this buffer system balances carbonic acids, bicarbonate ions and carbon dioxide. It serves to neutralises gastric in the stomach.
Assuming that the container is completely full, that the temperature is 22.0 ∘C, and that the atmospheric pressure is 1.2 atm , calculate the percent (by volume) of air that would be displaced if all of the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.)
Answer:
The % of displaced volume of nitrogen is 29.06%.
Explanation:
Volume of nitrogen = 1.2 L = 1200 mL
Density of nitrogen = 0.807 g/ml
Mass of nitrogen = [tex]Density \times Volume= 0.807 \times 1200 = 1044 g[/tex]
Molar mass of nitrogen = 28 g/mol
[tex]Number\,of\,moles\,nitrogen=\frac{Mass}{Molar\,mass}[/tex]
[tex]= \frac{1044}{28}=37.28[/tex]
The ideal gas equation is as follows
[tex]PV = nRT[/tex]
Rearrange the equation is as follows.
[tex]V= \frac{nRT}{P}...............(1)[/tex]
n= Number of moles = 37.28
R = Gas constant = 0.0820
T = Temperature = 22+ 273 = 295
P = Pressure = 1.2 atm
Substitute the all values in equation (1)
[tex]V= \frac{37.28 \times 0.0820 \times 295 }{1.2}= 751.5L= 0.751 \,m^{3}[/tex]
[tex]0.751 \,m^{3}[/tex] of nitrogen will displace same amount of air.
[tex]Volume \,\,of\,\, closet= 1 \times 1.3 \times 2 = 2.6\,m^{3}[/tex]
[tex]%\,displaced\,volume=\frac{0.751}{2.6}=28.8%[/tex]
Therefore, The % of displaced volume of nitrogen is 29.06%.
Final answer:
The percent by volume of air displaced by the evaporation of liquid nitrogen is calculated using the density of liquid nitrogen, the molar mass of nitrogen, and the Ideal Gas Law, accounting for the specific temperature and pressure conditions.
Explanation:
To calculate the percent by volume of air displaced if all the liquid nitrogen evaporated, we need to consider the Ideal Gas Law as well as the density of the liquid nitrogen. Since liquid nitrogen has a density of 0.807 g/mL, we can first find the mass of the nitrogen and then use the molar mass to determine the number of moles of nitrogen gas at the given temperature and pressure when it evaporates.
Using the Ideal Gas Law, PV = nRT, where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature in Kelvin. At standard temperature and pressure, the density of nitrogen gas can be calculated using the given molar mass and standard gas equation factors.
Once the volume of the gaseous nitrogen is determined, that volume will represent the amount of air volume displaced. To find the percent by volume displaced, divide the volume of gaseous nitrogen by the volume of the container and multiply by 100%.
Photophosphorylation produces ___ , which stores chemical energy that can be used for the Calvin cycle.
Photophosphorylation of sugar moiety produces ATP (adenosine triphosphate) which stores chemical energy that can be used for the Calvin cycle.
What is Calvin cycle?The Calvin cycle, often known as the Calvin-Benson cycle after its discoverers, is a series of chemical processes that collect carbon dioxide.
It is also referred to as the C3 cycle alone. C3 plants are defined as those that only fix carbon through the Calvin cycle. In the stroma of chloroplasts, carbon dioxide diffuses and mixes with the five-carbon sugar ribulose 1, 5-biphosphate (RuBP).
Rubisco, a large molecule that may be the most prevalent organic compound on Earth, is the name of the enzyme that catalyzes this reaction.
The three-carbon chemical 3-PGA is changed into another three-carbon compound termed G3P by ATP and NADPH using their stored energy. A reduction reaction is the name for this kind of reaction.
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Rank the following salts in order of decreasing pH of their 0.1 M aqueous solutions.(a) FeCl2, FeCl3, MgCl2, KClO2 .(b) NH4Br, NaBrO2, NaBr, NaClO2.
Answer:
a) FeCl2, FeCl3, MgCl2, KClO2.
KClO2 --> K+ + ClO2-; ClO2- will hydrolyse to form HClO +OH-
Mg+2, Fe+2 and Fe+3 ions will form acidic solutions, since theyfom slightly amount of
Mg+2 + 2H2O <-> Mg(OH)2 + 2H+
Fe+2 + 2H2O <-> Fe(OH)2 + 2H+
Fe+3 + 3H2O <-> Fe(OH)3 + 3H+
Therefore;
decreasing pH is high pH to low pH:
KClO2 > MgCl2 > FeCl2 > FeCl3
b) NH4Br, NaBrO2, NaBr, NaClO2.
NH4Br is acidic, forms NH4+ and NH4+ dnates H+ to form NH3 andH+
NaBrO2 is basic, forms Na+ + BrO2- then H2O + BrO2- HBrO2
NABr is neutral, NaClO2 is basics, forms Na+ + ClO2-then H2O + ClO2- HClO2
decreasig pH:
NaClO2 > NaBrO2 > NaBr >NH4Br
Note that HClO2 is stronger acid than HBrO2, therefore, expectmore HBrO2 formation
NaBrO2 > NaClO2 > NaBr >NH4Br
The order in (a) is; FeCl2, FeCl3, MgCl2, KClO2. The order in (b) is; NaBrO2 > NaClO2 > NaBr >NH4Br
The term pH refers to the degree of acidity or alkalinity of a solution. We must recall that salts are solvated in solution. The pH of the solution after solvation depends on the ions produced by the salt in solution.
Since FeCl2 yields a basic solution, then it has the highest pH. Similarly, KClO2 yields an acid solution hence it has the lowest pH. The order in (a) is; FeCl2, FeCl3, MgCl2, KClO2. The order in (b) is; NaBrO2 > NaClO2 > NaBr >NH4Br
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A food chemist determines the concentration of acetic acid in a sample of apple vinegar by acid base titration. The density of the sample is 1.01 g/mL. The titrant is 1.024 M NaOH. The average volume of titrant required to titrate 25.00 mL subsamples of the vinegar is 20.78 mL. What is the concentration of acetic acid in the vinegar?
Answer:
The concentration of acetic acid in the vinegar is 7,324 (%V/V)
Explanation:
The titration equation of acetic acid with NaOH is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
The moles required were:
1,024M×0,02500L = 0,02560 moles NaOH. These moles are equivalent (By the titration equation) to moles of CH₃COOH. As molar mass of CH₃COOH is 60,052g/mol, the mass in these moles of CH₃COOH is:
0,02560 moles CH₃COOH×[tex]\frac{60,052g}{1mol}[/tex]= 1,537g of CH₃COOH
As density is 1,01g/mL:
1,537g CH₃COOH×[tex]\frac{1mL}{1,01g}[/tex]= 1,522mL of CH₃COOH
As volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
[tex]\frac{1,522mLCH_{3}COOH}{20,78mL}[/tex]×100= 7,324 (%V/V)
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Considering the definition, molarity, density and volume percentage, the concentration of acetic acid in the vinegar is 7.32% (v/v).
The balanced reaction is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
Molarity is the number of moles of solute that are dissolved in a given volume. Molarity is determined by:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case you know for NaOH:
Molarity= 1.024 Mnumber of moles of solute= ?Volume= 25 mL= 0.025 LSo, the moles required are calculated as:
[tex]1.024 M=\frac{number of moles of solute}{0.025 L}[/tex]
Solving:
1.024 M× 0.025 L= number of moles of solute
0.0256 moles= number of moles of solute
By observing the balanced reaction, you can see that, since the ratio of CH₃COOH to NaOH is 1: 1, these moles are equivalent to the number of moles of CH₃COOH.
On the other hand, since the molar mass of CH₃COOH is 60.052 g/mol, this is the mass that contains one mole of the compound, the mass in 0.0256 moles of CH₃COOH is:
[tex]0.0256 molesx\frac{60.052 grams}{1 mole} =[/tex] 1.537 grams
Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
[tex]density=\frac{mass}{volume}[/tex]
In this case, being the density 1.01 [tex]\frac{g}{mL}[/tex], the volume can be calculated as:
[tex]1.01 \frac{g}{mL} =\frac{1.537 grams}{volume}[/tex]
Solving:
1.01 [tex]\frac{g}{mL}[/tex]× volume= 1.537 grams
[tex]volume=\frac{1.537 grams}{1.01 \frac{g}{mL} }[/tex]
volume= 1.522 mL of CH₃COOH
Volume Percentage (%v/v) is a measure of concentration that indicates the volume of solute per 100 volume units of the solution. In other words, the volume percent of a component in the solution is defined as the ratio of the volume of the component to the volume of the solution, expressed as a percentage.
The volume percentage of a solution is determined by the following expression:
[tex]volume percentage=\frac{volume of solute}{volume of solution}x100[/tex]
Then, as volume of vinegar in the sample is 20,78mL, the concentration of acetic acid in the vinegar is:
[tex]volume percentage=\frac{1.522 mL}{20.78 mL}x100[/tex]
Solving:
volume percentage= 7.32%
Finally, the concentration of acetic acid in the vinegar is 7.32% (v/v).
Learn more about:
molarity with this example: brainly.com/question/15406534?referrer=searchResultsdensity: brainly.com/question/952755?referrer=searchResults brainly.com/question/1462554?referrer=searchResultsvolume percentagebrainly.com/question/1713808If a chemical reaction such as photosynthesis begins with 6 atoms of carbon C , how many atoms of carbon C should be in the products?a. 12 atoms of carbon Cb. 6 atoms of carbon Cc. 3 atoms of carbon Cd. 2 atoms of carbon C
Answer:
Option B. 6 atoms of carbon
Explanation:
In any chemical reaction, atoms from reactant side and product side must be the same.
This is photosynthesis reaction:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
In the photosynthesis equation, 6CO₂ + 6H₂O becomes C₆H₁₂O₆ + 6O₂, conserving the number of carbon atoms. Therefore, if the reaction starts with 6 atoms of carbon, there will also be 6 atoms of carbon in the products.
When photosynthesis occurs, it follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Hence, the number of carbon atoms in the reactants must equal the number of carbon atoms in the products. The balanced chemical equation for photosynthesis is 6CO₂ + 6H₂O ⇒C₆H₁₂O₆ + 6O₂.
For this equation, 6 molecules of carbon dioxide (CO₂) containing 1 carbon atom each combine with water (H₂O) to produce glucose (C₆H₁₂O₆) and oxygen (O₂). Since we begin with 6 carbon atoms in the 6 molecules of carbon dioxide, the glucose product will also contain 6 carbon atoms because the carbon atoms are simply rearranged in this reaction.
Therefore, the answer to the question is: 6 atoms of carbon C should be in the products, which corresponds to option b. 6 atoms of carbon C.
The oxidative phase of the pentose phosphate pathway is___________.
A. reversible under physiological conditions.
B. has an enzyme requiring thiamine pyrophosphate as a coenzyme.
C. produces ribulose‑5‑phosphate and NADPH for biosynthetic processes.
D. converts pentoses to hexoses.
Answer:
C: Produces ribulose-5-phosphate and NADPH for biosynthetic processes.
Explanation:
There are two distinct phases in the pathway. The first is the oxidative phase, in which NADPH is generated, and the second is the non-oxidative synthesis of 5-carbon sugars. For most organisms, the pentose phosphate pathway takes place in the cytosol; in plants, most steps take place in plastids.
The oxidative phase of the pentose phosphate pathway generates ribulose-5-phosphate and NADPH for biosynthetic processes. Option C the correct choice.
The oxidative phase of the pentose phosphate pathway produces ribulose‑5‑phosphate and NADPH for biosynthetic processes. This phase starts with the oxidation of glucose-6-phosphate, catalyzed by glucose-6-phosphate dehydrogenase, generating NADPH and 6-phosphogluconolactone.
Following steps produce ribulose-5-phosphate, another molecule of NADPH, and release CO₂. Ribulose-5-phosphate can then be isomerized to ribose-5-phosphate or enter further reactions leading to the synthesis of other pentose phosphates as well as fructose 6-phosphate and glyceraldehyde 3-phosphate, which are intermediates in glycolysis.
Thus, the correct answer is C.
The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K. Calculate the standard change in enthalpy for this reaction (assuming ∆H° is temperature independent).
Answer:
[tex]\text { The standard change in enthalpy for this reaction is } \Delta \mathrm{H}=2.397 \times 10^{4} \mathrm{J} / \mathrm{mol}[/tex]
Explanation:
Let’s assume[tex]\text { At } 300 \mathrm{K}, \mathrm{k}_{\mathrm{eq}}=\mathrm{x}[/tex]
Thus as per given information
[tex]\text { At } 350 \mathrm{K}, \mathrm{k}_{\mathrm{eq}}=3.95 \mathrm{x}[/tex]
As we know:
[tex]\ln \left(\frac{k_{2}}{k_{1}}\right)=-\frac{\Delta \mathrm{H}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)[/tex]
[tex]\ln \left(\frac{3.95 x}{x}\right)=\frac{-\Delta \mathrm{H}}{8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}[/tex]
[tex]8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \times \ln 3.95=-\Delta \mathrm{H} \times\left(-4.761 \times 10^{-4} \mathrm{K}^{-1}\right)[/tex]
[tex]\Delta \mathrm{H}=\frac{8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} \times 1.373}{-4.761 \times 10^{-4} \mathrm{K}^{-1}}[/tex]
[tex]\Delta \mathrm{H}=2.397 \times 10^{4} \mathrm{J} / \mathrm{mol}[/tex]
The standard change in enthalpy for this reaction is mathematically given as
dH=2.397 *10^{4} J/mol
What is the standard change in enthalpy for this reaction?
Question Parameter(s):
The equilibrium constant for a certain reaction increases by a factor of 3.95 when the temperature is increased from 300.0 K to 350.0 K.
Generally, the equation for the change in enthalpy is mathematically given as
[tex]\ln \ (\frac{k_{2}}{k_{1}})=-\frac{\Delta \mathrm{H}}{R} (\frac{1}{T_{2}}-\frac{1}{T_{1}})[/tex]
Therefore
[tex]8.314 {J} {mol}^{-1} {K}^{-1} *\ln 3.95=-d{H} *(-4.761 \times 10^{-4} {K}^{-1})[/tex]
dH=2.397 *10^{4} J/mol
In conclusion
dH=2.397 *10^{4} J/mol
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The enthalpy of combustion of acetylene C2H2 is described by
C2H2 (g) + (5/2)O2 (g) >>>>>>>CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol
Calculate the enthalpy of formation of accetylene, given the following enthalpies of formation
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol
Answer: The enthalpy of formation of accetylene is, 226.2 kJ/mol
Explanation:
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The chemical equation for the combustion of acetylene follows:
(1) [tex]C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)[/tex]
[tex]\Delta H^o_{rxn}=-1299kJ/mol[/tex]
The formation of [tex]CO_2[/tex] will be,
(2) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=-393.5kJ/mol[/tex]
The formation of [tex]H_2O[/tex] will be,
(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
The formation of [tex]C_2H_2[/tex] will be,
(4) [tex]2C(s)+H_2(g)\rightarrow C_2H_2(g)[/tex] [tex]\Delta H_f_{(C_2H_2)}=?[/tex]
Now we are reversing equation 1, multiplying equation 2 by 2 and then adding equation 1, 2 and 3, we get:
Reaction (1) :
[tex]2CO_2(g)+H_2O(l)\rightarrow C_2H_2(g)+\frac{5}{2}O_2(g)[/tex]
[tex]\Delta H^o_{rxn}=1299kJ/mol[/tex]
Reaction (2) :
[tex]2C(s)+2O_2(g)\rightarrow 2CO_2(g)[/tex] [tex]\Delta H_f_{(CO_2)}=2\times -393.5kJ/mol=-787kJ/mol[/tex]
Reaction (3) :
[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_f_{(H_2O)}=-285.8kJ/mol[/tex]
[tex]\Delta H_f_{(C_2H_2)}=\Delta H^o_{rxn}+\Delta H_f_{(CO_2)}+\Delta H_f_{(H_2O)}[/tex]
[tex]\Delta H_f_{(C_2H_2)}=(1299kJ/mol)+(-787kJ/mol)+(-285.8kJ/mol)[/tex]
[tex]\Delta H_f_{(C_2H_2)}=226.2kJ/mol[/tex]
Therefore, the enthalpy of formation of accetylene is, 226.2 kJ/mol
To calculate the enthalpy of formation of acetylene, we can use the enthalpy change of the combustion reaction and the enthalpies of formation of CO2 and H2O. By rearranging the equation and applying Hess's law, we can determine the enthalpy of formation of acetylene.
Explanation:The enthalpy of formation of a compound can be calculated using Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken. To calculate the enthalpy of formation of acetylene (C2H2), we can use the equation:
C2H2 (g) + (5/2)O2 (g) → CO2 (g) + H2O (l)
The enthalpy change of this reaction is -1299 kJ/mol. By rearranging the equation, we can compare it to the standard formation equations for CO2 (g) and H2O (l) to determine the enthalpy of formation of acetylene.
The enthalpy of formation of acetylene is:
ΔHf(C2H2) = -1299 kJ/mol - [(-393.5 kJ/mol) + (-285.8 kJ/mol)]
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covalent molecular solids (like sugar) and covalent network solids (like diamond) have similar bonds and bond strengths. However, molecular solids are brittle and network solids are very strong and durable. Explain the difference.
Answer:
Covalent molecular solids (like sugar): No intermolecular covalent bonds
Covalent network solids (like diamond) : Intermolecular covalent interactions are present.
Explanation:
Covalent moleuclar solids are the solids which have "intera-molecular" covalent bonds i.e the covalent bonds exist between two atoms within the same molecule. The intermolecular bonds are not covalent. They are some weak intermolecular interactions like hydrogen bonds or disperison forces.
In case of covalent network solids, the intermolecular interactions are also strong covalent bonds. Thus they give a more strength and firmness to the compound or solid.
Heat capacity is a measure of the heat required to
A) expand the volume of a substance.
B) evaporate a substance.
C) freeze a substance.
D) raise the temperature of 1 gram of a substance by 1°C
Answer: Correct answer is D) Raise the temperature of 1 gram of a substance by 1 degree.
Explanation:
Heat capacity or thermal capacity is a physical property of matter, defined as the amount of heat to be supplied to a given mass of a material to produce a unit change in its temperature. This is rasi
When a radioactive isotope decays into a nucleus which is also unstable and undergoes decay, and this process is repeated several times, the succession of reactions is called a________.
a. fission reaction.
b. decay series.
c. half-life.
d. fusion reaction.
e. none of these
Answer:
Decay series
Explanation:
A succession of radioactive decay is termed decay series. The radios decay of an unstable nuclei usually continues until a stable isotope is reached. This continuous decay of radioactive isotopes is also known as a radioactive cascade.
It is important to note that most radioisotopes do not decay directly to form a stable nuclei. Instead, they undergo a series of decay until a stable isotope is formed. An example of a decay series can be seen in the decay of uranium-238 to uranium-234.
U-238 is more radioactive than U-234. U-238 first undergoes an alpha particle decay to form thorium 234. This is known as the daughter nuclei. Afterwards, thorium 234 undergoes decay to give protactinium 234. This then undergoes a beta negative decay to form the uranium 234 nuclei.
Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 3.49 g of butane is mixed with 7.0 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
The maximum mass of carbon dioxide that could be produced by the chemical reaction is 5.96 grams
Explanation:
The balanced reaction between gaseous butane and oxygen occurs as follows:
[tex]2 C_{4} H_{10} + 13 O_{2}[/tex]⇒[tex]8 CO_{2} +10 H_{2} O[/tex]
In order for the equation to be balanced, it was taken into account that the law of conservation of matter states that no atom can be created or destroyed in a chemical reaction, so the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
Knowing the reaction that occurs between both reagents, it is possible to know the stoichiometry of the reaction (that is, the quantities of reagents necessary for a certain amount of products to be produced). And assuming that 3.49 g of butane are mixed with 7.0 g of oxygen it is possible to determine the limiting reagent, that is to say the reagent that is consumed first, by determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
In order to determine the limiting reagent, you must first determine the reacting mass of each reagent. Then you must first know the molar mass of butane and oxygen, taking into account the atomic mass of each element that composes it and the amount present:
Atomic masses:
C: 12 g/molH: 1 g/molO: 16 g/molMolecular Mass:
[tex]C_{4}H_{10}[/tex]: 12 g/mol *4 + 1 g/mol *10= 58 g/molO₂: 16 g/mol *2= 32 g/molCO₂: 12 g/mol + 16 g/mol *2= 44 g/molH₂O: 1 g/mol *2 + 16 g/mol= 18 g/molTherefore, observing the reaction, 2 moles of butane and 13 moles of oxygen react. With the previously calculated molar masses it is possible to determine the mass that reacts by stoichiometry of each reagent.
Reactive mass of each reagent:
C₄H₁₀= 2 moles* 58 g/mol= 116 gO₂= 13 moles* 32 g/mol= 416 gAssuming that 3.49 g of butane react, and taking into account stoichiometry, it is possible to make a rule of three to determine the limiting reagent: if for 116 grams of butane to react, 416 grams of oxygen are needed, how many moles of oxygen are needed to 3.49 grams react of butane?
[tex]grams of oxygen=\frac{3.49 grams of butane* 416 grams of oxygen}{116 grams of butane}[/tex]
grams of oxygen= 12.52
Then the limiting reagent will be oxygen because a smaller amount of reagent (7 grams) is available. Then the following calculations are made from the available 7 grams of oxygen.
First, the amount of product that is produced by stoichiometry is determined, as was previously done with the reagents:
CO₂: 8 moles* 44 g/mol= 354 gH₂O: 10 moles* 18 g/mol= 180 gTo determine the maximum mass of carbon dioxide that could cause the chemical reaction, a rule of three is applied taking into account the limiting reagent and stoichiometry: if by stoichiometry 416 grams of oxygen produce 354 grams of carbon dioxide, how many grams of product produce 7 grams of oxygen?
[tex]mass of carbon dioxide=\frac{7 grams of oxygen* 354 grams of carbon dioxide}{416 grams of oxygen}[/tex]
mass of carbon dioxide= 5.96 grams
Finally, the maximum mass of carbon dioxide that could be produced by the chemical reaction is 5.96 grams
A sample of 0.0883 g of M g, which has a molar mass of 24.31 g/mol, produces 82.1 mL of H 2 gas. The gas is collected over water at atmospheric pressure of 766.7 mm Hg at 22 oC, at which the vapor pressure of water is 19.8 mm Hg. What is the experimental value for the molar volume of the gas in L/mol?
Answer:
24 Lt/mol
Explanation:
Though we have many data here (such as molar mass of Mg, water vapor pressure, etc), we need to focus on data for H₂, which will help us to obtain the molar volume of this gas
Statement refers the following data for H₂:
V = 82.1 ml = 0.082 Lt
T = 22°C = 295 K
P atm = 766.7 mm Hg = 1.0089 atm (which is, the pressure for H₂ before being collected in water)
If we consider H₂ behaves as an ideal gas:
PV = nRT
Then we can move some terms from this ecuation :
V/n = RT/P so we can obtain the relation between V (volume) and n (N° of moles) for H₂, which is the experimental valur for the molar volume
Considering R = 0.082 Lt*atm/K*mol:
V/n = [(0.082 Lt*atm/K*mol)x295 K]/1.0089 atm
V/n = 23.97 Lt/mol (molar volume at this experiment conditions)
The dissolution of 0.200 l of sulfur dioxide at 19 °c and 745 mmhg in water yields 500.0 ml of aqueous sulfurous acid. The solution is titrated with 13.4 ml of sodium hydroxide. What is the molarity of naoh?
Answer:
[tex]Molarity=1.22\ M[/tex]
Explanation:
Given:
Pressure = 745 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 745 / 760 = 0.9803 atm
Temperature = 19 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (19 + 273.15) K = 292.15 K
Volume = 0.200 L
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K
⇒n = 0.008174 moles
From the reaction shown below:-
[tex]H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O[/tex]
1 mole of [tex]H_2SO_4[/tex] react with 2 moles of [tex]NaOH[/tex]
0.008174 mole of [tex]H_2SO_4[/tex] react with 2*0.008174 moles of [tex]NaOH[/tex]
Moles of [tex]NaOH[/tex] = 0.016348 moles
Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)
So,
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.016348}{0.0134}\ M[/tex]
[tex]Molarity=1.22\ M[/tex]
Combustion Analysis of a compound containing N: While the composition of Carbon and Hydrogen can be determined directly from combustion analysis, elements like N and S have to be determined by separate methods. Use information provided for next 5 questions: A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of a 2.18 g sample burns in excess oxygen yields 3.94 g of CO2 and 1.89 g of H2O. A separate experiment shows that a 1.23 g sample contains 0.235 g of N. Calculate the moles of C in the sample.
Answer:
0.1 moles
Explanation:
Okay, the explanation here is pretty long. Let’s go!
Firstly, we know that the compound contains only carbon, hydrogen, oxygen and nitrogen. So the molecular formula would look like this CxHyOzNp
Where x , y, z and p are number of atoms of each element present respectively. According to the law of constant composition, the ratio of the number of atoms are fixed irrespective of the source or method of preparation. From this, we now know that in both samples, we have the same number of atoms. What is proper to do is to calculate the numbers, we do that as follows.
Firstly, we will need to calculate these numbers using the masses given in the first sample. Over the calculations, we should note that the formula we would be using is the relation: mass = number of moles * atomic mass or molar mass. Rearranging the equation gives different variations of the formula.
Now let’s do some mathematics.
There is 3.94g of carbon iv oxide, we can calculate the number of moles of it present which eventually would yield the number of moles of carbon present.
The molar mass of carbon iv oxide is 44g/mol.
The number of moles of carbon iv oxide present is thus 3.94/44 = 0.0895 moles
Since there is just 1 atom of carbon present in carbon iv oxide, this means the number of moles of carbon present is also 0.0895 moles
The mass of carbon present is the number of moles of it present multiplied by the atomic mass unit of carbon which is 12. This mass is 0.0895 * 12 = 1.0745g
Next, we calculate the number of moles of hydrogen and consequently its mass present.
To get this, we can access it from the number of moles of water present.
We get this by dividing the mass of water present by the molar mass of water. This is equals 1.89/18 = 0.105 moles. Now we know that 1 mole of water has 2 atoms of hydrogen, hence 1 mole of water will yield 2 moles of hydrogen. The number of moles of hydrogen present is thus 0.105 * 2 = 0.21 moles.
The mass of hydrogen thus present is 0.21 * 1 = 0.21g
Now, we know the mass of hydrogen present and the mass of carbon present. But, we do not know the mass of oxygen and nitrogen present.
To get this, we subtract the mass of hydrogen and carbon present from the mass of the total= 2.18 - 1.0745 - 0.21 = 0.8955g
Now we know the mass of oxygen and nitrogen combined. We can access their number of moles using their respective atomic masses. The total number of moles present is equal 0.8955/30 = 0.02985 moles
Wondering where 30 came from? The atomic mass of nitrogen and oxygen are 14 and 16 respectively. We now get the number of moles of both present. This is equal to the atomic mass divided by the total mass multiplied by the number of moles.
For oxygen = 16/30 * 0.02985 = 0.01592
For nitrogen = 14/30 * 0.02985 = 0.01393
From this we can try and get an empirical formula for the compound. This helps us to know the ratios of the number of atoms. To get this , we divide the number of moles by the smallest number of moles. The smallest number of moles is unarguably that of nitrogen. The empirical formula is calculated as follows:
C = 0.0895/0.01393 = 6
H = 0.21/0.01393 = 15
O = 0.01592/0.01393 = 1
N = 0.01393/0.01393 = 1
Thus, the empirical formula looks like this :
C6H15NO
Now, we can move to the second sample.
We know that the sample contains 0.235g of nitrogen. We first need to get the number of moles of nitrogen present in the sample. To get this, we simply divide this mass by the atomic mass. That is: 0.235/14 = 0.0168 moles
Now since the question asks us to get the number of moles of carbon and we know that in any elemental analysis, the ratio of carbon to nitrogen is 6 to 1, we simply multiply the number of moles of nitrogen by 6.
Hence, this is 0.0168 * 6 = 0.1 moles
Final answer:
The moles of carbon in the sample can be calculated using the mass of CO2 produced during combustion. The calculated moles of carbon are 0.0895 mol, based on the provided mass of CO2 and its molar mass.
Explanation:
The question asks how to calculate the moles of carbon in a sample based on combustion analysis data. To find the moles of carbon, we use the mass of CO2 produced during combustion. Since each mole of CO2 contains one mole of carbon, the moles of CO2 will equal the moles of carbon. The mass of CO2 produced is 3.94 g. The molar mass of CO2 is 44.01 g/mol.
Using the formula moles = mass / molar mass, the moles of carbon can be calculated as:
Moles of C = 3.94 g CO2 / 44.01 g/mol = 0.0895 mol C
This calculation determines the amount of carbon present in the sample by analyzing the carbon dioxide produced during combustion.
A client has been caring for an ailing mother for 18 months. In addition, the client has recently lost a job and is having difficulty finding a new one. Blood pressure has recently been running higher than normal. What is the client experiencing? 1 Allostasis 2 Homeostasis 3 Allostatic overload 4 Chronic homeostasis
Answer: The client is experiencing Allostatic overload.
Explanation:
Allostatic load is known as the wear and tear on the body which accumulates as an individual is exposed to repeated or chronic stress. "Chronic Stress" here, is caused by the difficulty of the client getting a new job coupled with the client caring for their ailing mother.
Which notation designates a possible value for a principal energy level in an atom?
px
2
+1/2
s
Answer:
2
Explanation:
There are four quantum numbers:
Principal quantum number (n)
Azimuthal quantum number (l)
Magnetic quantum number (ml)
Principal quantum number (n)
It tell about the energy levels. It is designated by n.
For example,
If n =2
It means there are two energy level present.
Azimuthal quantum number (l)
The azimuthal quantum number describe the shape of orbitals. Its value for s, p, d, f... are 0, 1, 2, 3. For l=3
(n-1)
4-1 = 3
it means principle quantum is 4 and electron is present in f subshell.
Magnetic quantum number (ml)
It describe the orientation of orbitals. Its values are -l to +l. For l=3 the ml will be -3 -2 -1 0 +1 +2 +3.
Spin quantum number (ms)
The spin quantum number tells the spin of electron either its clock wise (+1/2) or anti clock wise (-1/2).
If the electron is added in full empty orbital its spin will be +1/2 because it occupy full empty. If electron is already present and another electron is added then its spin will be -1/2.
A particular container holds 4.67 mol of neon gas. The volume of this container can be altered by sliding a piston in or out. The volume is changed from 8.90 L to 7.60 L while at the same time the temperature is changed from 344 K to 294 K. The molar heat capacity, CV,m, for neon is 12.47 J/(mol · K). Assume that this value will not change over the given temperature range. What is the change in entropy for the gas?
Answer:
15.24 J/K
Explanation:
First, let's write the expression to calculate the change in entropy:
ΔS = n*C*ln(T2/T1)
C is the heat capacity. It will be Cp if the heating is isobaric, and Cv if it is isochoric. (In this case is Cv)
Now, in order to do this, we need to calculate first the pressure of the gas at first, and then, after the chance:
P = nRT/V
P1 = 4.67 *¨0.082 * 344 / 8.9 = 14.801 atm
P2 = 4.67 * 0.082 * 294 / 7.6 = 14.814 atm
Now, let's calculate the change in entropy:
ΔS = 4.67 * (12.47 + 8.314) ln(344/294)
ΔS = 15.24 J/K
The vapor pressure of liquid antimony is 400 mm Hg at 1.84×103 K. Assuming that its molar heat of vaporization is constant at 115 kJ/mol, the vapor pressure of liquid Sb is 394.98 mm Hg at a temperature of 1.81×103 K. Find the vapor pressure of liquid Sb?
Answer:
Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg
Explanation:
The vapor pressure can be calculated by using Clausius‐Clapeyron equation.
ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)
Where
p₁ is the vapor pressure at T₁ (Initial Temperature)
p₂ is the vapor pressure at T₂ (final Temperature)
ΔHvap is molar heat of vaporization of the substance
R is the real gas constant = 8.314 x 10⁻³ kJ/mol.K
Data Given:
p₁ = ?
p₂ = 394.98 mm Hg
T₁ = 1.84×10³ K
T₂ = 1.81×10³ K
ΔHvap = 115 kJ/mol
Put the values in the Clausius‐Clapeyron equation
ln(p₁/p₂) = (-ΔHvap/R)(1/T₁ - 1/T₂)
ln(p₁/394.98 mm Hg) = (-115 kJ/mol / 8.314 x 10⁻³ kJ/mol.K)(1/1.84×10³ K- 1/1.81×10³ K)
ln(p₁ /394.98 mm Hg) = (- 13.8321 x 10³)(-0.5519)
ln(p₁ /394.98 mm Hg) = 7633.936
ln cancel out by E, e is raise to a power x
So,
p₁/394.98 mm Hg = e^7633.936
p₁/ 394.98 mm Hg = 20.75 x 10³
p₁ = 20.75 x 10³ x 394.98 mm Hg
p₁ = 8.19 x 10⁴ mm Hg
Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg
Answer:
338 mm Hg
Explanation:
[tex]ln(\frac{p2}{400mmHg} )=-115*\frac{1}{0.008314} *(\frac{1}{1.8*10^3} -\frac{1}{1.84*10^3}[/tex])
ln(p2/400)=-0.16705
p2/400= e^-0.16705=0.84615
p2=0.84615*400=338
Note that T2 is lower than T1 and that the vapor pressure decreases as the temperature decreases.
An apparatus consists of a 3 L flask containing nitrogen gas at 17◦C and 822 kPa, joined by a valve to a 10 L flask containing argon gas at 17◦C and 59 kPa. The valve is opened and the gases mix. a) What is the partial pressure of nitrogen after mixing?
b) What is the partial pressure of argon aftermixing?
c) What is the total pressure of the gas mixture?
Answer:
a) 189.7 kPa
b) 45.4 kPa
c) 235 kPa
Explanation:
After the mixture, the temperature must remain the same, because both gases are at 17°C. Besides, the gases will not react because both nitrogen and argon are non-reactive. So, the partial pressure can be calculated by Boyle's Law:
P1*V1 = P2*V2
Where P is the pressure, V is the volume, 1 is the initial state, and 2 the final state.
When the valve is opened, the gases mix and occupy all the two vases. So, V2 = 13 L.
a) For nitrogen, P1 = 822 kPa, V1 = 3L, V2 = 13 L
822*3 = P2*13
P2 = 189.7 kPa
b) For argon, P1 = 59 kPa, V1 = 10 L, V2 = 13 L
59*10 = P2*13
P2 = 45.4 kPa
c) By Dalton's law, the total pressure of a gas mixture is the sum of the partial pressures of the components, so:
P = 189.7 + 45.4
P = 235 kPa
Which of the following is TRUE? a) A neutral solution contains [H2O] = [H3O⁺] b) A neutral solution does not contain any H3O+ or OH- c) An acidic solution has [H3O⁺] > [OH⁻] d) A basic solution does not contain H3O+ e) None of the above are true.
Explanation:
pH is the negative logarithm of hydronium ion concentration present in a solution.
If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9. The solution has more [tex][H_3O^+][/tex] than [tex][OH^-][/tex]
If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14. The solution has more [tex][OH^-][/tex] than [tex][H_3O^+][/tex]
The solution having pH equal to 7 is termed as neutral solution.The solution has equal concentration of [tex][H_3O^+][/tex] and [tex][OH^-][/tex]
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What acid and what base would react in aqueous solution so that the following salts appear as products in the molecular equation? Write the balanced molecular equation for each reaction.
a. potassium perchlorate
b. cesium nitrate
c. calcium iodide
Answer:
A. Perchloric acid and potassium hydroxide
B. Trioxonitrate V acid and Cesium hydroxide
C. Hydroiodic acid and calcium hydroxide
Explanation:
An acid reacts with a base to form salt and water. The acid and the base that reacted for each of the compounds have been written in the answer above. The reaction equations are given below:
A. HClO4 (aq) + KOH (aq) ———> KClO4 (aq) + H2O (l)
B. HNO3 (aq) + CsOH (aq) ———-> CsNO3 (aq) + H2O (l)
C. 2HI (aq) + Ca(OH)2 (aq) ———> CaI2 (aq) + 2H2O (l)
The salts potassium perchlorate, cesium nitrate and calcium iodide are formed from the acids perchloric acid, nitric acid and hydroiodic acid respectively reacting with the bases potassium hydroxide, cesium hydroxide and calcium hydroxide. The reactions are balanced accordingly to ensure mass conservation.
Explanation:To form the given salts through an acid-base reaction, we can make use of the fact that a salt is formed when an acid and a base react together. For each salt, we would need a metal hydroxide (base) and an acid whose anion corresponds to the anion in the salt.
For potassium perchlorate (KClO4), the base would be potassium hydroxide (KOH) and the acid would be perchloric acid (HClO4). The balanced molecular equation would be: KOH + HClO4 → KClO4 + H2O. For cesium nitrate (CsNO3), the base would be cesium hydroxide (CsOH) and the acid would be nitric acid (HNO3). The balanced molecular equation would be: CsOH + HNO3 → CsNO3 + H2O. For calcium iodide (CaI2), the base would be calcium hydroxide (Ca(OH)2) and the acid would be hydroiodic acid (HI). The balanced molecular equation would be: Ca(OH)2 + 2HI → CaI2 + 2H2O.Learn more about Forming Salts from Acids and Bases here:
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Calculate the molarity and mole fraction of acetone in a 1.17 m solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.
Molarity and mole fraction of acetone in solution can be calculated by using the densities and molality. Molarity is derived by dividing number of moles of acetone by volume of solution and mole fraction is derived by dividing moles of acetone by total moles in solution.
Explanation:To calculate the molarity and the mole fraction of acetone in a 1.17 m solution of acetone (CH3COCH3) in ethanol (C2H5OH), we need to first convert the molal concentration (moles of solute per kilogram of solvent) to molar concentration (moles of solute per liter of solution).
Using given data, density of acetone is 0.788 g/cm3 and ethanol 0.789 g/cm3. As volumes are additive, volume of solution = volume of solute + volume of solvent. With this data, we can find number of moles of acetone and ethanol respectively.
For molarity (M), we divide number of moles of solute (acetone) by volume of solution in liters. For mole fraction (χ), we divide number of moles of solute (acetone) by total number of moles in solution (acetone + ethanol moles).
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Redox reactions can be written as two half-reactions, focusing on the gain or loss of electrons by one of the chemical substances. One half-reaction shows the oxidation while the other shows the reduction. When the two half-reactions are combined, the overall reaction is obtained.
Answer:
K2Cr2O7 + HI + HClO4 → Cr(ClO4)3 + KClO4 + I2 + H2O
↓
8HClO4 + K2Cr2O7 + 6HI → 3I₂ + 2Cr(ClO4)3 + 2KClO4 + 7H2O
Explanation:
I put and example of what you said.
Potassium dichromate and iodide acid react with perchloric acid to generate chromium perchlorate, potassium chlorate, iodine and water.
First of all think all the oxidation number of each element. One ox. number will increase and the other decrease, so those will be our half reactions (one of reduction, the other of oxidation).
In the case above, I in HI acts with -1 and I2 has 0 (all elements in ground state has 0 as oxidation number). (Increase) - Oxidation
Cr in K2Cr2O7 acts with +6, in Cr(ClO4)3 is 3+ (Decrease) - Reduction
So, the first half reaction is:
2I⁻ → I₂ + 2e⁻ (OXIDATION)
I have to put 2 iodides to ballance, so the total charge is 2-. It has to release 2 electrons.
Cr2O7²⁻ → 2Cr³⁺
In products side, I have to add 2 chromes but we don't have the charges ballanced. At the main equation, we have acids so this redox occurs in an acidic medium. In the acidic medium we add water, the same as oxygen we have, so:
Cr2O7²⁻ → 2Cr³⁺ + 7H2O
and in reactant side, we add protons the same as hydrogen, we have, in this case like this
14H⁺ + Cr2O7²⁻ → 2Cr³⁺ + 7H2O
finally, we add the electrons. Chrome to decrease +6 to +3 had to lose 3 electrons but, we have 2 Cr, so 6 in total. These are final the 2 half reaction
14H⁺ + Cr2O7²⁻ + 6e- → 2Cr³⁺ + 7H2O (REDUCTION)
2I⁻ → I₂ + 2e⁻ (OXIDATION)
Electrons are not ballanced, we have to multiply by a minimum common multiple. For 2 and 6, this number is 12 so:
(14H⁺ + Cr2O7²⁻ + 6e- → 2Cr³⁺ + 7H2O) .2
(2I⁻ → I₂ + 2e⁻) .6
Afterwards, we can sum the reactions:
28H⁺ + 2Cr2O7²⁻ + 12e- + 12I⁻ → 6I₂ + 12e⁻ + 4Cr³⁺ + 14H2O
As we have 12e- in both sides, we cancel them
28H⁺ + 2Cr2O7²⁻ + 12I⁻ → 6I₂ + 4Cr³⁺ + 14H2O (still balanced)
Look that all the stoichiometry is even, so we can /2.
14H⁺ + Cr2O7²⁻ + 6I⁻ → 3I₂ + 2Cr³⁺ + 7H2O
So the final reaction is:
8HClO4 + K2Cr2O7 + 6HI → 3I₂ + 2Cr(ClO4)3 + 2KClO4 + 7H2O
We have in total 14H+, so 6 protons are for HI and 8 for the HClO4.
How much heat is absorbed when 30.00 g of C(s) reacts in the presence of excess SO2(g) to produce CS2(l) and CO(g) according to the following chemical equation?
5 C(s) + 2 SO2(g) → CS2(l) + 4 CO(g) ΔH° = +239.9 kJ
The reaction of 30 g of C(s) in the given equation absorbs approximately 119.95 kJ of heat.
Explanation:In chemistry, the amount of heat absorbed or released during a chemical reaction is given by the product of the mole ratio and the enthalpy change (ΔH°). Looking at the balanced equation: 5 C(s) + 2 SO2(g) → CS2(l) + 4 CO(g) ΔH° = +239.9 kJ, the reaction of 5 moles of C(s) absorbs 239.9 kJ of heat. However, we only have 30 g of C(s) - or 30/12.01 (since the molar mass of carbon is 12.01 g/mol) equals roughly 2.5 mol of C(s).
To find out how much heat 2.5 mol of C(s) absorbs, we use a proportional relationship: (2.5 moles C/5 moles C) * 239.9 kJ = 119.95 kJ. Therefore, when 30 g of C(s) reacts in the presence of excess SO2(g) to produce CS2(l) and CO(g), it absorbs approximately 119.95 kJ of heat.
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The heat absorbed when 30.00 g of C(s) reacts in the given chemical reaction is approximately 119.95 kJ. This value is determined based on the moles of carbon involved and the enthalpy change given for the reaction.
First, we need to find the number of moles of carbon (C) involved in the reaction:
Molar mass of C = 12.01 g/molMoles of C = 30.00 g / 12.01 g/mol ≈ 2.50 molesThe balanced chemical equation shows that 5 moles of C are needed for one reaction cycle, which absorbs 239.9 kJ of heat.
To find the heat absorbed for 2.50 moles of C, we use the proportion:
Heat absorbed = (2.50 moles / 5 moles) × 239.9 kJ ≈ 119.95 kJ
Therefore, the heat absorbed when 30.00 g of C(s) reacts is ≈ 119.95 kJ.
Which is an example of transforming potential energy to kinetic energy? Select two options. changing thermal energy to electrical energy changing chemical energy to thermal energy changing nuclear energy to radiant energy changing radiant energy to electrical energy changing mechanical energy to chemical energy
Answer:
changing chemical energy to thermal energy
changing nuclear energy to radiant energy
Explanation:
Chemical energy can also be referred to as chemical potential energy which is the energy stored in the bonds of chemical compounds while thermal energy is a form of kinetic energy because it also involves the movement of particles.
Nuclear energy is the potential energy present in an atom while radiant energy is a form of kinetic energy that involves the movement of particles or waves during electromagnetic radiation.
The examples of transforming potential energy to kinetic energy are changing the chemical energy in food to kinetic energy for movement, such as riding a bicycle, and converting electromagnetic radiation (light energy) from the sun into chemical energy through photosynthesis in plants.
Explanation:Transforming potential energy to kinetic energy occurs when an object, initially at rest due to its position in a gravitational field, descends. As the object falls, potential energy decreases, and kinetic energy increases. This conversion follows the principle of conservation of energy, with the total energy remaining constant, neglecting external factors like air resistance.
The examples of transforming potential energy to kinetic energy are: changing the chemical energy in food to kinetic energy for movement, such as riding a bicycle, and converting electromagnetic radiation (light energy) from the sun into chemical energy through photosynthesis in plants.
A container holds a mixture of two gases at The partial pressures of gas A and gas B, respectively, are and If of a third gas is added with no change in volume or temperature, what will the total pressure become?
Answer:
The total pressure will be the sum of the partial pressure of these three gases.
Explanation:
According to Dalton law of partial pressure,
The total pressure exerted by mixture of gases is equal to the sum of partial pressure of the individual gas.
This expression can be written as,
P(total) = P₁ + P₂ + P₃ + ...... Pₙ
For example:
If the pressure of A is 2 atm and Partial pressure of B is 4 atm the total pressure will be,
P(total) = P₁ + P₂
P(total) = 2 atm + 4atm
P(total) = 6 atm
when third gas is added which exert the partial pressure of 4 atm,
Then total pressure will becomes,
P(total) = P₁ + P₂ + P₃
P(total) = 2 atm + 4atm + 4 atm
P(total) = 10 atm
The reaction is found to be first order in IO3−, first order in SO32−, and first order in H+. If [IO3−]=x, [SO32−]=y, and [H+]=z, what is the rate law for the reaction in terms of x, y, and z and the rate constant k?
Answer:
rate = kxyz
Explanation:
It is worth knowing that the rate low can only be determined by experimentation only not by just balancing equations. So here we are told that all the reactants x , y and z are all first order. This is important because we use this as exponents. That is why the exponents of all the reactants will be 1.
rate = kxyz