Answer:
Cellular respiration
Explanation:
Oxygen is required by the cell in the Krebs cycle as the ultimate proton acceptor (creating water an end product during making out ATPs). During the conversion of pyruvate from glycolysis to Acetyl-CoA that enters the Krebs cycle, the pyruvate is decarboxylated (or oxidized) hence creating carbon IV oxide as a byproduct.
What mass of co2 is produced by the combustion of 1.00 mol of ch4?
Final answer:
The combustion of 1.00 mol of CH4 produces 44.01 grams of CO2, as one mole of methane combusts to create one mole of carbon dioxide, and the molar mass of CO2 is 44.01 g/mol.
Explanation:
The mass of CO2 produced by the combustion of 1.00 mol of CH4 can be calculated by looking at the balanced chemical equation:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
From the equation, it is clear that one mole of methane (CH4) produces one mole of carbon dioxide (CO2) upon combustion. To convert moles of CO2 to grams, we use the molar mass of CO2, which is 44.01 g/mol.
Calculation:
1.00 mol CH4 × (44.01 g CO2 / 1 mol CO2) = 44.01 g CO2
Therefore, the combustion of 1.00 mol of CH4 will produce 44.01 grams of CO2.
How many grams of fluorine must be reacted with excess lithium iodide to produce 10.0 grams of lithium fluoride?
Calculate the theoretical yield if 100.0 g p4o10 react with 200.0 g h2o
Know how the rate of a chemical reaction is defined in terms of reactants or products.
How are genes related to DNA?
A-Genes are sections of DNA that program the amino acid sequences of proteins.
B-Genes are monomers that are joined by dehydration synthesis to make DNA.
C-Genes are proteins that bind molecules of DNA together to make amino acids.
D-Genes are monomers that are joined by hydrolysis to make DNA.
Solid metals undergo greater thermal expansion than liquids do.
Answer:
false
Explanation:
Final answer:
Thermal expansion is the increase or decrease in size of a body due to a change in temperature. Solid metals undergo greater thermal expansion than liquids. This is because the closely packed atoms or molecules in solid metals are pushed farther apart by the increase in temperature, resulting in a larger size for the whole body.
Explanation:
Thermal expansion: Thermal expansion is the increase, or decrease, of the size (length, area, or volume) of a body due to a change in temperature. It occurs in all dimensions - length, area, volume - and is not limited to solid metals. However, solid metals undergo greater thermal expansion than liquids do. The increased thermal expansion in solid metals can be attributed to their closely packed atoms or molecules, which are pushed farther apart by the increase in temperature, resulting in a larger size for the whole body.
How many nitrogen molecules are in a 8.86 L container of nitrogen gas at a pressure of 112.1 kPa and a temperature of 30.8 oC?
Potassium-40 has a half-life of 1.3 billion years. as the potassium-40 isotope decays, it becomes argon. if a rock was formed with 12 g of potassium-40, approximately how long would it take for 75% of the potassium-40 to be replaced by argon?
a. 1.3 billion years
b. 2.6 billion years
c. 5.2 billion years
d. 650 million years
Final answer:
It takes two half-lives, or 2.6 billion years, for 75% of Potassium-40 to decay to Argon, making (b) 2.6 billion years the correct option.
Explanation:
The decay of Potassium-40 (K-40) to Argon (Ar-40) follows the principles of radioactive decay and the concept of half-lives. The half-life is the time it takes for half of the original quantity of a radioactive isotope to decay. In this case, K-40 has a half-life of approximately 1.3 billion years. After one half-life, there would be 6 grams of K-40 and 6 grams of Ar-40. After two half-lives, which is 2.6 billion years, there would be 3 grams of K-40 and 9 grams of Ar-40. Since the question asks for the time it would take for 75% of the potassium-40 to be replaced by argon, we are looking for the time it takes to have only 25% of the original K-40 remaining.
Starting with 12 grams of K-40, after the first half-life, we would be left with 6 grams of K-40, which is 50%. After the second half-life, we would be left with 3 grams of K-40, which is 25% of the original amount. Therefore, it would take two half-lives for 75% of the potassium-40 to decay to argon, which equates to 2.6 billion years. The correct option is therefore (b) 2.6 billion years.
You have 1329.0 g of a bleach solution. the percent by mass of the solute sodium hypochlorite, naocl, is 3.62%. how many grams of naocl are in the solution? g
What is the theoretical yield in liters of hydrogen at STP if 20.1g of calcium are completely reacted
How much is 2.50 g of CuCl2 in moles ?
An iv solution contains 5.10g glucose in 100.5 ml water. what is the molarity of the solution? the molar mass of glucose is 180g/mol. molarity: mol solute/l solution
The molarity of the solution is 0.279 M.
Explanation:To find the molarity of the solution, we need to calculate the number of moles of glucose and the volume of the solution in liters. The molarity is defined as moles of solute per liter of solution.
First, we calculate the number of moles of glucose:
Moles of glucose = mass of glucose / molar mass of glucose
Moles of glucose = 5.10 g / 180 g/mol = 0.028 moles
Next, we calculate the volume of the solution in liters:
Volume of solution = 100.5 ml / 1000 = 0.1005 L
Finally, we plug in the values into the molarity formula:
Molarity = moles of glucose / volume of solution
Molarity = 0.028 moles / 0.1005 L = 0.279 M
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How does nh3 and ca(OH)2 form OH- ions in water?
A gas sample has a volume of 150 ml at 0.00 ∘c. the temperature is raised (at constant pressure) until the volume is 218 ml. what is the temperature of the gas sample in ∘c at this volume?
How many moles of gas occupy 98 l at a pressure of 2.8 atmospheres and a temperature of 292 k?
"11.46" moles of gas occupy 98 L at 2.8 atm pressure.
Ideal gas equation:The equation characterizing the stages of hypothetical gasses stated mathematically involving combinations of empirical as well as physiological constants, is considered as Ideal gas equation.
According to the question,
Volume occupied, V = 98
Temperature, T = 292 k
Pressure, P = 2.8 atm
By using ideal gas equation, we get
→ PV = nRT
or,
→ n = [tex]\frac{PV}{RT}[/tex]
By substituting the above values, we get
= [tex]\frac{2.8\times 98}{0.082\times 292}[/tex]
= [tex]\frac{274.4}{23.944}[/tex]
= [tex]11.46[/tex]
Thus the above answer is appropriate.
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The two naturally occuring isotopes of chlorine are 35cl and 37cl. based on the average atomic mass of cl (35.5 amu), what is the approximate natural abundance of 35cl?
The half-life of the radioactive element unobtanium dash 53 is 20 seconds. if 32 grams of unobtanium dash 53 are initially present, how many grams are present after 20 seconds? 40 seconds? 60 seconds? 80 seconds? 100 seconds?
Answer: After 20 seconds, 16 grams of unobtanium will be present, after 40 seconds, 8 grams of unobtanium will be present after 60 seconds, 4 grams of unobtanium will be present, after 80 seconds, 2 grams of unobtanium will be present and after 100 seconds, 1 grams of unobtanium will be present.
Explanation: Half life is the time in which half of the reaction is completed. Thus the half of the substance will be decomposed and half of it will remain.
Amount of the substance left after n half lives will be=[tex]\frac{A}{2^n}[/tex]
where A= initial amount of substance
n=no of half lives=[tex]\frac{\text{given time}}{\text{half life}}[/tex]
a) t= 20 seconds
no of half lives=[tex]\frac{20}{20}=1[/tex]
amount of the substance left after 1 half life will be=[tex]\frac{32}{2^1}[/tex]=16 g.
b) t= 40 seconds
no of half lives=[tex]\frac{40}{20}=2[/tex]
amount of the substance left after 2 half lives will be=[tex]\frac{32}{2^2}[/tex]=8g.
c) t= 60 seconds
no of half lives=[tex]\frac{60}{20}=3[/tex]
amount of the substance left after 3 half lives will be=[tex]\frac{32}{2^3}=4g[/tex]
d) t= 80 seconds
no of half lives=[tex]\frac{80}{20}=4[/tex]
amount of the substance left after 4 half lives will be=[tex]\frac{32}{2^4}=2g[/tex]
e) t= 100 seconds
no of half lives=[tex]\frac{100}{20}=5[/tex]
amount of the substance left after 5 half lives will be=[tex]\frac{32}{2^5}=1g.[/tex]
The half-life of a radioactive element is defined as the time required by the specific isotope to decrease by half of its original value. The unobtanium after given time will be present as:
After 20 secs, 16 gramsAfter 40 secs, 8 gramsAfter 60 secs, 4 gramsAfter 80 secs, 2 gramsAfter 100 secs, 1 gramHalf-life is the time required by the unobtanium is the half of the reaction is completed. Half of the substance will be decomposed, such that:
Amount of substance left n half-lives = [tex]\dfrac{\text A}{{2}^{\text n}}[/tex]
where A= initial amount of substance
Now,
n = number of half-lives = given time /half-life
Given,
1. Time = t = 20 seconds
Number of half-lives = [tex]\dfrac{20}{20}[/tex] = 1Amount of the substance left after 1 half life = [tex]\dfrac{\text 32}{{2}^{\text 1}}[/tex] = 16 grams.2.Time = t = 40 seconds
Number of half-lives = [tex]\dfrac{40}{20}[/tex] = 1Amount of the substance left after 1 half life = [tex]\dfrac{\text 32}{{2}^{\text 2}}[/tex] = 8 grams.3.Time = t = 60 seconds
Number of half-lives = [tex]\dfrac{60}{20}[/tex] = 1Amount of the substance left after 1 half life = [tex]\dfrac{\text 32}{{2}^{\text 3}}[/tex] = 4 grams.4.Time = t = 80 seconds
Number of half-lives = [tex]\dfrac{80}{20}[/tex] = 1Amount of the substance left after 1 half life = [tex]\dfrac{\text 32}{{2}^{\text 4}}[/tex] = 2 grams.5.Time = t = 100 seconds
Number of half-lives = [tex]\dfrac{100}{20}[/tex] = 1Amount of the substance left after 1 half life = [tex]\dfrac{\text 32}{{2}^{\text 5}}[/tex] = 1 grams.To know more about half-life, refer to the following link:
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What is the standard cell notation for an electrolytic cell with aluminum and gold electrodes?
Answer:
Au3+
Explanation:
The standard cell notation for an electrolytic cell with aluminum and gold electrodes Au(s)| Au³⁺ || Al³⁺| Al (s)
What is an Electrolytic Cell ?A device in which electrical energy is converted into chemical energy or chemical energy is converted into electric energy is called an Electrolytic cell
The electrolytic cell consists of two metallic electrodes and an electrolyte.
In this cell nomenclature, the electrode to the left of the salt bridge is always assumed to be the anode, and the accompanying half-equation is always stated as an oxidation , while right side is cathode and the half equation is Reduction.
Reaction at Anode: Au(s) → Au³⁺(aq) + 3e⁻ ..........(oxidation)
Reaction at Cathode: Au³⁺(aq) + 3e⁻ →Au(s) ..............(reduction)
The standard cell notation for an electrolytic cell with aluminum and gold electrodes
Au(s)| Au³⁺ || Al³⁺| Al (s)
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What is a characteristic of all fuel cells?
Energy from the Sun is converted into electricity.
Electrical energy is produced from oxidation reactions.
Oxidation reactions are used to produce thermal energy.
A constant source of energy is not required.
Answer:
Electrical energy is produced from oxidation reactions.
Explanation:
In fuel cells elecric energy is produced from the oxidation of reactants, that fuel is often Hydrogen, that is storaged in the fuel cell and then it just grabs oxygen from the air and converts that chemical reaction of oxidation in electric current.
Rubbing alcohol is 70.% isopropyl alcohol by volume. how many ml of isopropyl alcohol are in a 1 pint (473 ml) container? 330 ml 680 ml 0.15 ml 70. ml 470 ml
A 94.7-g sample of silver (s = 0.237 j/(g · °c)), initially at 348.25°c, is added to an insulated vessel containing 143.6 g of water (s = 4.18 j/(g · °c)), initially at 13.97°c. at equilibrium, the final temperature of the metal–water mixture is 22.63°c. how much heat was absorbed by the water? the heat capacity of the vessel is 0.244 kj/°c.
Q = m*C*dT
m = mass
C = specific heat capacity
dT = Temperature change
Q = heat evolved
Heat lost by metal is equal to the sum of heat taken up by water and vessel. Since water is taken in the vessel so
the initial and final temperatures would be same for vessel and water...
Q = 94.7 *0.237 *(22.63 - 348.25) = 7308.18 Joules = 7.3 KJ
Heat lost by silver = 7.3 KJ
Heat absorbed by water = 143.6*4.18*(22.63 - 13.97) = 5.2 KJ
The half-life for beta decay of strontium-90 is 28.8 years. a milk sample is found to contain 10.3 ppm strontium-90. how many years would pass before the strontium-90 concentration would drop to 1.0 ppm?
Answer : The correct answer is 96.68 yrs
Radioactivity Decay :
it is a process in which a nucleus of unstable atom emit energy in form of radiations like alpha particle , beta particle etc .
Radioactive decay follows first order kinetics , so its rate , rate constant , amount o isotopes can be calculated using first order equations .
The first order equation for radioactive decay can be expressed as :
[tex] ln \frac{N}{N_0} = - k*t [/tex] ----------- equation (1)
Where : N = amount of radioisotope after time "t"
N₀ = Initial amount of radioisotope
k = decay constant and t = time
Following steps can be used to find time :
1) To find deacy constant :
Decay constant can be calculated using half life . Decay constant and half life can be related as :
[tex] T _\frac{1}{2} = \frac{ln2}{k} [/tex] ---------equation (2)
Given : Half life of Strontium -90 = 28.8 years
Plugging value of [tex] T_\frac{1}{2} [/tex] in above formula (equation 2) :
[tex] 28.8 yrs = \frac{ln 2}{ k } [/tex]
Multiply both side by k
[tex] 28.8 yrs * k = \frac{ln 2 }{k} * k [/tex]
Dividing both side by 28.8 yrs
[tex] \frac{28.8 yrs * k}{28.8 yrs} = \frac{ln 2}{28.8 yrs} [/tex]
(ln 2 = 0.693 )
k = 0.0241 yrs⁻¹
Step 2 : To find time :
Given : N₀ = 10.3 ppm N = 1.0 ppm k = 0.0241 yrs⁻¹
Plugging these value in equation (1) as :
[tex] ln (\frac{1.0 ppm}{10.3 ppm} ) = - 0.0241 yrs^-^1 * t [/tex]
[tex] ln (0.0971 ) = -0.0241 yrs ^-^1 * t [/tex]
(ln 0.0971 = - 2.33 )
Dividing both side by - 0.0241 yrs⁻¹
[tex] \frac{-2.33}{-0.0241 yrs^-^1} = \frac{-0.0241 yrs^-^1 * t}{-0.0241 yrs^-^1} [/tex]
t = 96.68 yrs
Hence the concentration of Strontium-90 will drop from 10.3 ppm to 1.0 ppm is 96.68 yrs
96.9 years
Further explanationGiven:
The half-life for beta decay of strontium-90 is 28.8 years. A milk sample is found to contain 10.3 ppm strontium-90.Question:
How many years would pass before the strontium-90 concentration would drop to 1.0 ppm?
The Process:
In the calculations of half-lives, the expressions are the following:
[tex]\boxed{ \ N = \frac{N_o}{2^n} \ }[/tex]
where [tex]\boxed{ \ n = \frac{t}{t_{1/2}} \ }[/tex] are used. In these expressions;
N₀ = initial number = 10.3 ppmN = amount of substances remained = 1.0 ppmt = time passedn = the number of half-lives[tex]t_{1/2} =[/tex] half-live = 28.8 yearsStep-1: find out the number of half-lives (n)
[tex]\boxed{ \ N = \frac{N_o}{2^n} \ } \rightarrow \boxed{ \ 2^n = \frac{N_o}{N} \ }[/tex]
[tex]\boxed{ \ 2^n = \frac{10.3}{1.0} \ }[/tex]
[tex]\boxed{ \ 2^n = 10.3 \ }[/tex]
[tex]\boxed{ \ n \cdot \ln{2} = \ln{10.3} \ } \rightarrow \boxed{ \ n = \frac{\ln{10.3}}{\ln{2}} \ }[/tex]
We get n = 3.364572
Step-2: find out in how many years would pass before the strontium-90 concentration would drop to 1.0 ppm
[tex]\boxed{ \ n = \frac{t}{t_{1/2}} \ } \rightarrow \boxed{ \ t = n \times t_{1/2} \ }[/tex]
t = 3.364572 x 28.8
t = 96.9
Thus in 96.9 years will pass before the concentration of strontium-90 will drop to 1.0 ppm.
- - - - - - - - - -
Notes:
The half-life of radioactive decay is the period of time required for half of the initial amount of the substance to disintegrate. The shorter the half-life of radioactive decay, the higher the rate of radioactive decay and the more radioactivity. The half-life is the characteristic property of each element.
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the scattering of light by particles in a mixture
Answer:The correct answer is option:'the scattering of light by particles in a mixture'.
Explanation:
Tyndall effect is defined as scattering of the light by the particles present in the mixture. A beam of the light passing through these mixture are visible to the human eyes. Generally colloidal mixture displays this type of effect. The name of his effect is named after the physicist named John Tyndall.
Hence, the correct answer is option:'the scattering of light by particles in a mixture'.
The term used to describe the rapid release of bubbles from a liquid is _____
How many electrons are transferred during the formation of one molecule of al2o3?
In the formation of one molecule of aluminum oxide (Al2O3), a total of 6 electrons are transferred from two aluminum atoms to three oxygen atoms to maintain electrical neutrality.
Explanation:The formation of one molecule of aluminum oxide (Al2O3) involves the transfer of electrons between aluminum atoms and oxygen atoms. Each aluminum atom loses three electrons, becoming Al3+ ions, while each oxygen atom gains two electrons to become O2- ions. To maintain electrical neutrality, the compound must have equal numbers of positive and negative charges.
Therefore, to form one molecule of Al2O3, two aluminum atoms (2 Al) will lose a total of 6 electrons (2 Al × 3 e-), and three oxygen atoms will gain a total of 6 electrons (3 O × 2 e-) during the electron transfer process. The complete transfer involves 12 electrons changing places but, in effect, the number of electrons transferred is the total on one side of the reaction which is 6.
J. J. Thomson’s experiment disproved the theory that an atom is divisible. is indivisible. contains protons. contains electrons.
J.J. Thomson's experiment did not disprove the theory that atoms are indivisible. Instead, his experiment demonstrated the presence of subatomic particles, specifically electrons, within atoms.
The atom's indivisibleness was not refuted by J. J. Thomson's experiment. Instead, his research demonstrated that electrons, a type of subatomic particle, are present in atoms. The "plum pudding" model of the atom was created as a result of Thomson's research from the late 19th and early 20th centuries, which advanced knowledge of atomic structure.
Thomson studied the behavior of electrically charged particles (electrons) inside a vacuum tube in his well-known cathode ray tube experiment. A stream of negatively charged particles (electrons) traveled from the cathode (negative electrode) to the anode (positive electrode) when a voltage was applied across the tube. Thomson came to the conclusion that these particles were the basic building blocks of atoms.
The old belief that atoms were indestructible was refuted by Thomson's discovery of electrons within them, which also supported the notion of subatomic particles inside the atom. His "plum pudding" concept proposed that atoms had a neutral overall charge because electrons were contained within a positively charged "pudding" or matrix.
Later investigations, like those by Ernest Rutherford, helped to clarify our understanding of atomic structure and revealed the existence of a positively charged nucleus that contains protons as well as neutral particles known as neutrons. The existence of indivisible atoms was therefore not refuted by Thomson's experiment, but it did introduce the idea of subatomic particles within atoms, particularly electrons.
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What do microarrays allow scientists to do
How many sites on this antibody molecule have potential to bind to a non-self molecule?
What is the general word equation for salt formation by neutralization?
metal + acid → salt + water
nonmetal + acid → salt + water
acid + base → salt + water
metal + base → salt + water
Answer: acid + base → salt + water
Explanation: Neutralization is a type of double displacement reaction in which exchange of ions take place. Neutralization is a chemical reaction between an acid and a base where ions exchange and lead to formation of salt and water.
[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O[/tex]
HCl is an acid which donates [tex]H^+[/tex] ions when dissolved in water. [tex]NaOH[/tex] is base which donates [tex]OH^-[/tex] ions when dissolved in water. They combine to form [tex]NaCl[/tex] which is salt and [tex]H^+[/tex] and [tex]OH^-[/tex] ions combine to form water [tex]H_2O[/tex].
A 2.5% (by mass) solution concentration signifies that there is of solute in every 100 g of solution. 2. therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be . 3. a solution mass of 1 kg is times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain of ki.