Answer:
The answer to your question is Ionic
Explanation:
Data
Phosphorous and Lithium
type of bond = ?
Process
In chemistry there are 2 main bonds:
Ionic is a bond between a metal and a nonmetal but also we can identify these bonds if the difference of electronegativity is higher than 1.7.
Covalent is a bond between two nonmetals or when the difference of electronegativity is lower than 1.7.
For this molecule:
Lithium is metal and Phosphorous is a nonmetal so the bond is Ionic.
or
Electronegativity for Lithium = 0.98
Electronegativity for Phosphorous = 2.19
Difference of electronegativity = 1.21
As 1.21 > 1.7 the bond is Ionic.
A low level of carbon dioxide labeled with 14C is accidentally released into the atmosphere surrounding industrial workers as they resume work following the lunch hour. Unknowingly, they breathe the contaminated air for 1 hour. Which of the following compounds will be radioactively labeled?
A) About one half of the carbon atoms of newly synthesized fatty acids.
B) All of the carbon atoms of newly synthesized fatty acid.
C) The carboxyl atom of newly synthesized fatty acids.
D) One half of the carbon atoms of newly synthesized acetyl CoA.
E) About one third of the carbons of newly synthesized malonyl CoA.
Answer:
D) One half of the carbon atoms of newly synthesized acetyl CoA.
Explanation:
It will be radioactively labeled because Malonyl CoA which contains 3 Carbon molecule is synthesized from Acetyl CoA which has 2 Carbon molecule.
This happens with the addition of ‘CO2’ with the help of the enzyme called acetyl CoA carboxylase.
Final answer:
The compounds that will become radioactively labeled after inhaling 14C-labeled carbon dioxide are the carboxyl atoms of newly synthesized fatty acids, as this is the only portion derived directly from CO₂. Thus, option C is correct.
Explanation:
To determine which compounds will become radioactively labeled after industrial workers breathe air containing 14C-labeled carbon dioxide, we must understand the metabolic pathways that involve the incorporation of carbon from CO₂. During the synthesis of fatty acids, acetyl CoA serves as the two-carbon donor in the form of its activated methyl group, while malonyl CoA provides a two-carbon unit that loses one carbon as CO₂ during the elongation cycle. Though both are involved, individual carbon atoms from CO₂ do not directly become part of the fatty acid chain.
Given that the carboxyl group of the fatty acids does not originate from acetyl CoA or malonyl CoA, the correct choice would be C) The carboxyl atom of newly synthesized fatty acids. This is because the carboxyl group is the only portion derived from the original CO₂ breathed by the workers.
The atomic mass of europium is 151.9 amu. There are two naturally occuring isotopes on europium, eu-151 with a mass of 151.0 amu and eu153 with a mass of 153.0. The percent natural abundance of the eu151 should be
Answer:
55%
Explanation:
Let A represent isotope eu-151
Let B represent isotope eu-153
Let A% represent Abundance of isotope A (eu-151)
Let B represent abundance of isotope B (eu-153)
The abundance of isotope A (eu-151) can be obtained as follow:
Step 1:
Data obtained from the question include:
Atomic mass of Europium = 151.9 amu
Mass of isotope A (eu-151) = 151.0 amu
Mass of isotope B (eu-153) = 153.0
amu
Abundance of isotope A (eu-151) = A%
Abundance of B (eu-153) = B% = 100 - A%
Step 2:
Determination of the abundance of Abundance of isotope A (eu-151). This is illustrated below:
Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]
151.9 = [(151 x A%)/100] + [(153x B%)/100]
151.9 = [(151 x A%)/100] + [(153x (100-A%))/100]
151.9 = [151A%/100] + [15300 -
153A%/100]
151.9 = (151A% + 15300 - 153A%) /100
Cross multiply to express in linear form
151.9 x100 = 151A% + 15300 - 153A%
15190 = 151A% + 15300 - 153A%
Collect like terms
15190 - 15300= 151A% - 153A%
- 110 = - 2A%
Divide both side by - 2
A% = - 110 / - 2
A% = 55%
Therefore the abundance of eu-151 is 55%
Answer:
The natural abundance of eu-153 is 45.0 %
The natural abundance of eu-151 is 55.0 %
Explanation:
Step 1: Data given
The atomic mass of europium is 151.9 amu.
eu-151 hass with a mass of 151.0 amu
eu153 hass a mass of 153.0 amu
Step 2: Calculate the percent natural abundance
natural abundance eu-151 = X
natural abundance eu-153 = 1-X
151.9 = 151.0* X + 153.0 * (1-X)
151.9 = 151.0X + 153.0 -153.0 X
-1.1 = -2.0 X
X = 0.55 = 55 %
The natural abundance of eu-153 is 45.0 %
The natural abundance of eu-151 is 55.0 %
A tiger cub has a pattern of stripes on its fur that is similar to that of its parents. Where are the instructions stored that provide information for a tiger’s fur pattern?
Answer:
On genes within chromosomes
Explanation:
-The fur pattern is a genetic trait whose instruction is carried in the DNA's base sequence.
-These instructions are organized into genes and each gene is only responsble for the making of a functional product.
-Hence, the instructions are stored on genes within chromosomes.
A branch of science which deals with a living being is called biology.
The correct answer is DNA.
The DNA stands for Deoxyribonucleic acid. it contains all the information about the parents.
The transfer of character from one generation to another is called inheritance.
In this case which is mentioned in the question, the stripes character is a transfer from the parent to the offspring in the form of a gamete cell. Gamete has all the information of the parent which shows the character after maturing.
Hence, the correct answer is DNA stored in the gamete.
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It takes 86 joules of energy to raise a box 2 meters using the pulley setup shown above.
Without the pulley, the box can be raised 2 meters using only 82 joules of energy. What could explain this 4-joule difference?
A.
When raising the box using the pulley, 4 joules of mechanical energy are broken down into protons, neutrons, and electrons.
B.
When raising the box using the pulley, 4 joules of mechanical energy are transformed into water.
C.
When raising the box using the pulley, 4 joules of mechanical energy are converted into heat energy due to friction.
D.
When raising the box using the pulley, 4 joules of mechanical energy are destroyed.
Answer:
the correct answer is c
Answer:
The answer is C, when raising the box using the pulley, 4 joules of mechanical energy are converted into heat energy due to friction. :)
Explanation:
Find the empirical formula of a compound found to contain 26.56% potassium, 35.41% chromium and the remainder oxygen.
Please help ASAP!!!
Answer: [tex]KCrO_4[/tex]
Explanation:
First, calculate how much percent of oxygen there is. We know that the whole compound cannot exceed 100%, so we take that, and substract it from 26.56% and 35.41%.
100.00
- 26.56
-------------------
73.44
- 35.41
---------------------
38.03
That is how much of oxygen we have.
Potassium: 26.56% or 26.56g
Chromium: 35.41% or 35.41g
Oxygen: 38.03% or 38.03g
To find the empirical formula, you simply find the amount of mol that each one has. You can do this by using the atomic mass of each element.
[tex]Potassium: 26.56g(\frac{1mol}{39.10g})= 0.67mol[/tex]
[tex]Chromium: 35.41g(\frac{1mol}{52g})=0.68mol[/tex]
[tex]Oxygen: 38.03g(\frac{1mol}{16g})=2.38mol[/tex]
We now determine the lowest number and divide each mol by it. In this case, the lowest number is 0.67
[tex]Potassium: \frac{0.67mol}{0.67}=1[/tex]
[tex]Chromium:\frac{0.68mol}{0.67}=1[/tex]
[tex]Oxygen:\frac{2.38mol}{0.67}=3.55 = 4[/tex]
Finally, we take each element add add their respective number.
So, this empirical formula would be:
[tex]KCrO_4[/tex]
The empirical formula of a compound with 26.56% potassium, 35.41% chromium, and the rest being oxygen would be [tex]K_2Cr_2O_7[/tex]
The number of mole of each element in the compound can be found by dividing each element's percentage with their respective molar weights:
Potassium K = 26.56%
= 26.56/39.1
= 0.68
Chromium, Cr = 35.41%
= 35.41/52
= 0.68
Oxygen, O = 100 - 26.56+35.41
= 38.03/16
= 2.38
Divide each number of moles by the smallest.
K = 0.68/0.68
= 1
Cr = 0.68/0.68
= 1
O = 2.38/0.68
= 3.5
Thus, the empirical formula would be [tex]KCrO_{3.5}[/tex]
Multiply all by 2 to remove the fraction:
[tex]K_2Cr_2O_7[/tex]
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Examine this reaction:
starch⟶amylasedisaccharides⟶disaccharidasemonosaccharides
If there is the normal amount of amylase present but less disaccharidase than usual, which of the following would most likely happen?
No starch could be broken down to disaccharides.
There would be less disaccharide and less monosaccharide produced.
The normal amount of disaccharide would be produced, but fewer monosaccharides would be produced.
There would be no overall impact because the normal amount of amylase is present.
Answer:
The normal amount of disaccharide would be produced, but fewer monosaccharides would be produced.
Explanation:
The first reaction, the conversion of starch into disaccharides, is catalyzed by the enzyme amylase. Since amylase is present in a normal amount, a normal amount of disaccharides will be produced.
In the second reaction, these disaccharides will be transformed into monosaccharides by a disaccharidase. However, since there is less disaccharidase, there will be fewer monosaccharides produced than if it was a normal amount of amylase.
Final answer:
If there is a normal amount of amylase but less disaccharidase than usual, the normal amount of disaccharide would be produced, but fewer monosaccharides would result from the reaction.
Explanation:
When examining the reaction starch → amylase → disaccharides → disaccharidase → monosaccharides, the presence of a normal amount of amylase means that starch can be broken down into disaccharides effectively. However, with less disaccharidase than usual, fewer monosaccharides would be produced from the disaccharides. The normal amount of disaccharide produced by amylase would not change, but since disaccharidase is responsible for the breakdown of disaccharides into monosaccharides, a deficiency in this enzyme means the reaction would be incomplete, leading to less monosaccharides being produced.
A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6 hours at 30°C. Assuming the kinetics of the microorganisms enzymatic action is responsible for the rate of decay, what is the activation energy for the decomposition process? Hint: Rate varies INVERSELY with time; a faster rate produces a shorter decomposition time. 1.67.2 kJ/mol 2.2.34 kJ/mol 3.23.4 kJ/mol 4.0.45 kJ/mol
Answer:
1. 67.2 kJ/mol
Explanation:
Using the derived expression from Arrhenius Equation
[tex]In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})[/tex]
Given that:
time [tex]t_1[/tex] = 8.3 days = (8.3 × 24 ) hours = 199.2 hours
time [tex]t_2[/tex] = 10.6 hours
Temperature [tex]T_1[/tex] = 0° C = (0+273 )K = 273 K
Temperature [tex]T_2[/tex] = 30° C = (30+ 273) = 303 K
Rate = 8.314 J / mol
Since [tex](\frac{k_2}{k_1}=\frac{t_2}{t_1})[/tex]
Then we can rewrite the above expression as:
[tex]In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})[/tex]
[tex]In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})[/tex]
[tex]2.934 = \frac{E_a}{8.314}(\frac{30}{82719})[/tex]
[tex]2.934 = \frac{30E_a}{687725.766}[/tex]
[tex]30E_a = 2.934 *687725.766[/tex]
[tex]E_a = \frac{2.934 *687725.766}{30}[/tex]
[tex]E_a =67255.58 \ J/mol[/tex]
[tex]E_a =67.2 \ kJ/mol[/tex]
Plasma spray-coating processes are often used to provide surface protection for materials exposed to hostile environments, which induce degradation through factors such as wear, corrosion, or outright thermal failure. True or false?
Final answer:
It is true that plasma spray-coating processes are used to protect materials from environments which could cause degradation due to factors like wear, corrosion, or thermal failure. Corrosion, a REDOX process, can be combatted with protective coatings, including more nobel metals or cathodic protection via sacrificial anodes.
Explanation:
True, plasma spray-coating processes are indeed utilized to provide surface protection for materials that are exposed to hostile environments. These environments may induce degradation through wear, corrosion, or thermal failure. Corrosion, specifically, is a REDOX (reduction-oxidation) process where metals deteriorate through oxidation. This fact is evident in how iron rusts or how copper develops a patina when exposed to air. Protective coatings are a common method to prevent corrosion, as they can consist of a second metal that is more difficult to oxidize, or contain more easily oxidized metals, providing cathodic protection. For instance, the application of a thin layer of zinc can protect galvanized steel and sacrificial electrodes can be attached to protect metals such as iron.
Which conditions contain molecules with the same average kinetic energy as the molecules in 10.0 grams of co2 at 10 c?
Answer:
40g of CO2 at 10C
Explanation:
Since average kinetic energy depends on absolute temperature (directly proportional to absolute temperature) and independent of amount and nature of gas. given that is have same temperature.
40g of CO2 at 10C
Any gas at 10 ℃ would have the same average kinetic energy as 10.0 grams of CO2 at the same temperature due to the principles of the kinetic molecular theory.
The conditions that contain molecules with the same average kinetic energy as the molecules in 10.0 grams of CO2 at 10 ℃ would be any other mass of gas at the same temperature, since the average kinetic energy of the molecules of a gas depends only on temperature according to the kinetic molecular theory. This is true regardless of the type of gas or its mass, as long as the gases are at the same temperature.
For example, if you have helium gas at 10 ℃, the average kinetic energy of its molecules would be the same as that of the CO2 molecules at 10 ℃. This is because the kinetic molecular theory posits that all gases at the same temperature have the same average kinetic energy.
Explore: use the gizmo to explore whether the number of atoms present affects the half-life that you measure. describe your findings below: _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________ _________________________________________________________________________
Answer:
The number of atoms present does not affects the half-life
Explanation:
The half-life, [tex]t_{1/2}[/tex] of an unstable atomic nuclei is the duration in time for the quantity of the atoms having such unstable nuclear to decrease by a factor of 2 in nature. Here nature refers to the position of the unstable nuclear in the environment. It is indicative of the rate of decay of unstable atoms and the period of survival of stable atoms
The half life is given by
[tex]N(t) = N_0 \left (\frac{1}{2} \right )^{\frac{t}{t_{1/2}}}[/tex]
Therefore,
[tex]t_{1/2} = \frac{t\cdot ln2}{ln(\frac{N_0}{N_t} )}[/tex]
Whereby as time time increases N₀ becomes larger such that [tex]t_{1/2}[/tex] remain constant.
Rock and mineral fragments come from ____________________________ and are small particles of sediment such as ________________________________________________.
Answer:
Rocks can be simply defined as the aggregates of various minerals, and these broken rock fragments and mineral grains are formed from the weathering of rocks. Weathering refers to the breakdown of rocks due to the occurrence of various geological processes that are initiated by the agents such as wind, water and ice.
The weathering process leads to the formation of sediments of variable size and shape, and these are such as pebbles, sand, silt and clay particles.
Which solution is the most concentrated?
you have a picture so i can see it
The empirical formula/s mass is equal to or greater than that of the molecular formula’s molecular mass.
Answer:
The empirical formula is the representation of simplest whole number ratios of the elements which comprising the compound while, molecular formula is the actual whole number ratio of elements in the formation of compounds.
Empirical formula is equal to molecular formula.
Calculate delta s ,delta s total when the volume of 123 g CO initially at 298K and 1 bar increased by a factor of four in (a) an adiabatic reversible expansion
Final answer:
The change in entropy (ΔS) for an ideal gas during a reversible adiabatic process is zero because no heat is exchanged with the surroundings. Hence, the total change in entropy (ΔS_total) is also zero for such a process.
Explanation:
To calculate the change in entropy (ΔS) of 123 g of carbon monoxide (CO) during an adiabatic reversible expansion, one should use the thermodynamic definition of entropy. However, since no specific heat is provided for CO, and the problem implies an ideal gas behavior, usually entropy change for an ideal gas in an adiabatic process is zero because no heat is transferred into or out of the system (ΔQ = 0). In adiabatic processes, all expansion work is done at the expense of the internal energy of the gas, and since entropy is a function of heat exchange at a given temperature, the net change in entropy for the gas in such a process is zero.
For the second part of the question, computing the total entropy change ΔS_total during the expansion would typically require considering the surroundings as well. In a truly adiabatic and reversible expansion, there is no heat exchange with the surroundings, thus no change in the entropy of the surroundings, and the total entropy change would likewise be zero. In the real world or in irreversible processes, ΔS_total might be different, but these require additional information.
describe how revolution and year are related.
Dissolving potassium chlorate (KClO3) is even more endothermic than potassium chloride.
Your task is to determine how many grams of potassium chlorate you would have to add to 100 mL of water to produce the same temperature change as 5 grams of KCl.
Answer:
The mass of KClO₃ that will absorb the same heat as 5 g of KCl is 3.424 g
Explanation:
Here we have
Heat of solution of KClO₃ = + 41.38 kJ/mol.
Heat of solution of KCl (+17.24 kJ/mol)
Therefore, 1 mole of KCl absorbs +17.24 kJ during dissolution
Molar mass of KCl = 74.5513 g/mol
Molar mass of KClO₃ = 122.55 g/mol
74.5513 g of KCl absorbs +17.24 kJ during dissolution, therefore, 5 g will absorb
[tex]\frac{17.24}{74.5513 } \times 5 \, \, kJ \, or \, 1.156 \, kJ[/tex]
Therefore the amount of KClO₃ to be dissolved to absorb 1.156 kJ of energy is given by
122.55 g of KClO₃ absorbs + 41.38 kJ, therefore,
[tex]\frac{1.156}{41.38} \times 122.55 \, g = 3.424 \, g[/tex]
Therefore the mass of KClO₃ that will absorb the same heat as 5 g of KCl = 3.424 g.
If both potassium chlorate and potassium chloride had the same specific heat, you would need 5 grams of potassium chlorate to produce the same temperature change in 100 mL of water. This is a simplified scenario and in real-life, there would be slight differences.
Explanation:To answer this question, we need to assume that potassium chlorate and potassium chloride have the same specific heat and that the water's heat capacity is much greater than either salt. Therefore, the amount of heat energy required to increase the temperature of water is the same for both salts. If 5 grams of potassium chloride produce the same temperature change, an equal amount of heat energy would be required for the potassium chlorate. Hence, you would need 5 grams of potassium chlorate to produce the same temperature change.
Please understand that this is a simplified explanation. In a real-life situation, there would be some differences due to the different chemistry of the salts.
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Fluorine and Nitrogen react to form a covalent molecule. Which best describes the molecule they will make?
There are 3 Nitrogens and 1 Fluorine. Nitrogen will be the central atom.
There are 3 Fluorines and 1 Nitrogen. Fluorine will be the central atom.
There are 3 Nitrogens and 1 Fluorine. Fluorine will be the central atom.
There are 3 Fluorines and 1 Nitrogen. Nitrogen will be the central atom.
Final answer:
Fluorine and nitrogen react to form a molecule where nitrogen is the central atom, bonded to three fluorine atoms. This is due to nitrogen's ability to form multiple bonds and its relative electronegativity compared to fluorine.
Explanation:
The correct description of the molecule formed by fluorine and nitrogen is: There are 3 Fluorines and 1 Nitrogen. Nitrogen will be the central atom. In molecules where nitrogen is combined with halogens like fluorine, nitrogen often serves as the central atom due to its capacity to form multiple bonds and its comparably lower electronegativity than halogens. This is consistent with the general rule that less electronegative atoms tend to be central in molecular structures. For instance, nitrogen (N) has five valence electrons and can share these electrons to form covalent bonds, whereas fluorine (F), being more electronegative, typically forms single bonds and is often found as a terminal atom.
3500 J of energy are added to a 0.5 mol sample of iron at 293 K. What is the final temperature of the iron in kelvins? The molar heat of iron is 25.1 J/(mol•K).
Answer:
[tex]T_2=571.9K[/tex]
Explanation:
Hello,
In this case, we consider the following formula defining the energy and the temperature change for the sample of iron:
[tex]Q=n_{Fe}Cp_{Fe}(T_2-T_1)[/tex]
Now, solving the final temperature, considering a positive inlet heat, we have:
[tex]T_2=T_1+\frac{Q}{n_{Fe}Cp{Fe}} =293K+\frac{3500J}{0.5mol*25.1J/(mol*K)} \\T_2=571.9K[/tex]
Best regards.
Answer:
Explanation:
Step 1: Data given
Number of heat transfer = 3500 J
Number of moles of iron = 0.5 moles
Initial temperature = 293 K
The molar heat of iron is 25.1 J/(mol*K)
Step 2: Calculate ΔT
Q = n* C * ΔT
⇒with Q = the heat transfer = 3500 J of energy
⇒with n = the number of moles iron = 0.5 moles
⇒with C = the molar heat of iron = 25.1 J/mol*K
⇒ΔT = the change of temperature = T2 - T1 = T2 - 293 K
3500 J = 0.5 moles *25.1 J/mol * K * ΔT
ΔT = 278.9
Step 3: Calculate ΔT
ΔT = 278.9 = T2 - T1 = T2 - 293 K
T2 = 278.9 + 293 K
T2 = 551.9 K
Identify the orbitals that overlap to form the C−Cl bonds in CH2Cl2. Identify the orbitals that overlap to form the bonds in . carbon sp3 hybrid orbital with a singly occupied chlorine 3s orbital carbon sp3 hybrid orbital with chlorine sp3 hybrid orbital carbon sp2 hybrid orbital with a singly occupied chlorine 3p orbital a singly occupied carbon 2p orbital with chlorine sp3 hybrid orbital carbon sp3 hybrid orbital with a singly occupied chlorine 3p orbital
Complete Question
Identify the orbitals that overlap to form the C−Cl bonds in CH2Cl2.
a. carbon sp3 hybrid orbital with a singly occupied chlorine 3s orbital
b. carbon sp2 hybrid orbital with a singly occupied chlorine 3p orbital
c. carbon sp3 hybrid orbital with chlorine sp3 hybrid orbital
d. carbon sp3 hybrid orbital with a singly occupied chlorine 3p orbital
e. a singly occupied carbon 2p orbital with chlorine sp3 hybrid orbital
Answer:
Correct Option is D
Explanation:
The molecular formula of the compound given is
[tex]CH_2 Cl_2[/tex]
The structural formula for this given compound is shown on the first uploaded image
Looking at the structural formula we see that all the bonds are single bonds which shows that carbon is sp3 hybridzied which means that one 2s orbital of carbon has mixed with 3 2p orbital to for a form a four hybrid orbital as shown on the second uploaded image
For the clorine the outer shell is containing two 3s orbital which are completely filled and a 6 3p orbital which requires an electron to complete it as shown on the uploaed image
Hence the bond between the carbon and the clorine is between a
sp3 hybridzied orbital and a 3p orbital
Note: each orbital contains a single electron
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Joe is concerned about the carbon dioxide emissions from his daily commute and the impact that it is having on global warming In his daily drive to and from work Joe travels 60 miles and gets 20mpg. Joe would like to know how many grans if carbon dioxide is emitted from his driving each day.
Amount of CO₂ emission per day is 11,356.23 g.
Explanation:
Joe travelling distance per day = 60 miles
Carbon dioxide emission per day = 20 mpg
Now we have to find the amount of carbon dioxide emitted per day by dividing the distance by the emission per day given in gallons.
Amount of Carbon dioxide emission = [tex]$\frac{distance}{emission amount}[/tex]
Amount of CO₂ emission in gallons = [tex]$\frac{ 60 miles}{20 mpg}[/tex]
= 3 gallons
Now we have to convert the gallons to grams as,
1 gallon = 3,785.41 g
3 gallons = 3 × 3785.41 g = 11,356.23 g
So the emission of CO₂ per day is 11,356.23 g.
HELP!!
Which claim about how temperature relates to heat is correct?
A Temperature is a measure of the heat of an object.
B Heat causes a change in the temperature of an object.
C Raising the temperature causes the heat of an object to increase.
D Temperature and heat are two different ways to measure the same thing.
Answer:
raising the temperature increases heat in the object... heat is a form of energy that is induced in an object by raising temperature
C
Explanation:
raising temperature causes the heat of an object to increase
Under which circumstance might a gas decrease in volume when heated? A)The gas is held constant at STP. B)The gas remains under uniform pressure. C)tHE GAS IS PLACE UNDER INCREASING PRESSURE. D)The gas undergoes a decrease in pressure
Answer is option C. The gas is placed under increased pressure.
Explanation:
According to Gas laws,
If the temperature to gas increases then the molecules of gas gets additional energy and moves with more spped than earlier which gives more more colliding to the walls of container. This increases pressure.
The volume and gas in a container are directly proportional to each other.
This means if the volume gets decreased then the collision of gas molecules to the walls increases which results increase in pressure.
Therefore, volume is inversely proportional to pressure.
Select the correct answer.
Which term describes the unstable arrangement of the reacting particles as the reaction takes place?
A.
reactants
B.
activation energy
C.
products
D.
energy of reaction
E.
activated complex
Answer:
E. activated complex
Explanation:
If a helium balloon breaks loose, it rises into the atmosphere and at some point bursts. Why
Answer:
Explanation:By the time, that the balloon got too high in the sky, the pressure inside the balloon will soon overcome the pressure outside, and the balloon's elasticity is not too strong to hold the air inside, that the Helium gas inside will successfully push out the walls of the balloon and so it bursts!
How many moles of KBr will be produced from 7 moles of BaBry?
BaBr2+ K2SO4 → KBr +BaSO4
7 moles of BaBr₂ produces 14 moles of KBr.
Explanation:
Given reaction is
BaBr₂ + K₂SO₄ → KBr + BaSO₄
It can be balanced by putting 2 in front of KBr as,
BaBr₂ + K₂SO₄ → 2 KBr + BaSO₄
From the above balanced equation, it was found that, 1 mole of Barium bromide required to produce 2 moles of KBr.
Now the molar ratio of BaBr₂ to KBr is written as 1 : 2.
In the same way, the molar ratio for 7 moles of BaBr₂, can be written as 7:14.
So 7 moles of BaBr₂ produces 14 moles of KBr.
The moles of KBr formed has been 14 mol.
The balanced chemical equation for the reaction has been:
[tex]\rm BaBr_2\;+\;K_2SO_4\;\rightarrow\;2\;KBr\;+\;BaSO_4[/tex]
From the balanced chemical equation, the coefficient has been describing the moles of reactants and products formed.
The balanced chemical equation has been given as:
[tex]\rm 1\;mol\;BaBr_2=2\;mol\;KBr[/tex]
The moles of KBr formed by 7 moles of Barium bromide has been as:
[tex]\rm 1\;mol\;BaBr_2=2\;mol\;KBr\\7 \;mol\;BaBr_2=2\;\times\;7 \;mol\;KBr\\7 \;mol\;BaBr_2=14 \;mol\;KBr\\[/tex]
The moles of KBr formed has been 14 mol.
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The reaction below is spontaneous under standard conditions - true or false? br2(l) + mn2+(aq) → mno4-(aq) + br-(g) (in acidic solution) use the fact that the reduction potential for mno4-(aq) is +1.51 v and for br2(l) it is +1.065 v.
Answer:
The answer is false
Explanation:
The given reaction is
[tex]Br_{2}_{(l)} + Mn^{2+}_{(aq)} ------->MnO_4 _{(aq)} + Br^-_{(g)}[/tex]
At the cathod the potential is
[tex]E_{cath} = + 1.065V[/tex] [This because it will attract the negative
charge]
At the Anode the potential is
[tex]E_{anode} = 1.51[/tex]
The overall potential of the cell is
[tex]E_{cell} = E_{cath} - E_{anode}[/tex]
[tex]= -0.445[/tex]
Since the potential of the cell is less than 0 then the reaction is not spontenous under standard condition
The reaction below is spontaneous under standard conditions should be considered false.
Calculation of reaction:Since the given reaction is br2(l) + mn2+(aq) → mno4-(aq) + br-(g)
Also, at the anode, the potential should be
Eanode = 1.51
And, the overall potential of the cell is
= -0.445
Also, the potential of the cell should be lower than 0 so the reaction is not spontenous under standard condition
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5. A 28.4 g sample of aluminum is heated to 39.4 °C, and placed in a calorimeter containing 50.0 g of water. The
temperature of water increases from 21.00 °C to 23.00 °C. What is the specific heat capacity, C, of aluminum?
Answer:
C Al = 0.8975 J/g.K
Explanation:
Q = mCΔT∴ m Al = 28.4 g
∴ T Al = 39.4°C = 312.4 K
∴ m H2O = 50.0 g
∴ T1 H2O = 21°C = 294 K
∴ T2 H2O = 23°C = 296 K
∴ C H2O = 4,18 J/g.K
⇒ C Al = ?
in a calorimeter:
∴ Al give heat: Q Al < 0
∴ H2O revceives heat: Q H2O > 0
⇒ - Q Al = Q H2O
⇒ - (28.4 g)*(C Al)*(296 K - 312.4 K) = (50.0 g)*(4.18 J/g.K)*(296 K - 294 K)
⇒ - (- 465.76 g.K)*(C Al) = 418 J
⇒ C Al = (418 J) / (465.76 g.K)
⇒ C Al = 0.8975 J/g.K = 897.5 J/Kg.K
The specific heat capacity of aluminum in this example is 0.394 J/g °C. This was determined by calculating the heat gained by water and considering that it equals the heat lost by the aluminum, and subsequently solving for the specific heat capacity of aluminum.
Explanation:To find the specific heat capacity (C) of aluminum, we must consider the amount of heat transferred from the aluminum to the water (expressed as q). The heat gained by water is calculated using the equation q = m * C * ΔT, where m is the mass, C is the specific heat, and ΔT is the change in temperature. According to the question, water's mass (m) is 50.0g, its specific heat (C) is 4.184 J/g °C, and the difference in its temperature (ΔT) is 2.00 °C. So, the heat gained by water is q = 50.0g * 4.184 J/g * °C * 2.00 °C = 418.4 J.
The heat lost by aluminum is equal to the heat gained by water. Therefore, using the equation q = m * C * ΔT and plug in the values of q (418.4 J), m (28.4 g), and ΔT (39.4 °C), to solve for C, the specific heat of aluminum, we can rearrange the formula to C = q / (m * ΔT) = 418.4 J / (28.4g * 39.4 °C) = 0.394 J/g °C.
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Hydrogen sulfide gas reacts with oxygen gas to form sulfur dioxide and water. What volume of oxygen is needed to completely react with 2.8 Liters hydrogen sulfide?
Answer:
4.2 L O₂ is needed to completely react with 2.8 L hydrogen sulfied.
Explanation:
Without pressure and temperature we cannot calculate the this vale
We assume that the reaction take place under standard Temperature and Pressure(STP).
At STP, One mole ([tex]6.023\times 10^{23}[/tex] particles) of any gas occupied volume 22.4 L.
The balanced equation of this reaction is
[tex]2H_2S+3O_2\rightarrow 2SO_2+2H_2O[/tex]
Now we use molar ratio.
[tex]2.8L\ H_2S . \ \frac{1 mol\ H_2S}{22.4L \ H_2S}\ . \ \frac{3 mol\ O_2}{2 mol \ H_2S} \ . \ \frac{22.4L\ O_2}{1 mol\ O_2}[/tex]
=4.2 L O₂
4.2 L O₂ is needed to completely react with 2.8 L hydrogen sulfied.
Final answer:
To determine the volume of oxygen needed to react with 2.8 liters of hydrogen sulfide, the stoichiometric ratio from the chemical equation is used, revealing that 4.2 liters of oxygen is required for complete reaction.
Explanation:
Stoichiometry of Hydrogen Sulfide and Oxygen Reaction
The question involves the stoichiometric relationship between hydrogen sulfide and oxygen gases during a chemical reaction. The balanced chemical equation for the reaction between hydrogen sulfide (H₂S) and oxygen (O₂) to produce sulfur dioxide (SO₂) and water (H₂O) is:
2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)
To find the volume of oxygen needed to react with 2.8 liters of hydrogen sulfide, we use the stoichiometric coefficients from the balanced equation, which tell us that 2 volumes of H₂S react with 3 volumes of O₂.
This gives us a ratio of:
2H₂S : 3O₂
Using this ratio, we can find the volume of oxygen required by setting up the proportion:
(2.8 L H₂S) / (2L H₂S) = (x L O₂) / (3L O₂)
Solving for x, we get:
x = (2.8 L H₂S) × (3L O₂) / (2L H₂S)
x = 4.2 L O₂
Therefore, 4.2 liters of oxygen gas is needed to completely react with 2.8 liters of hydrogen sulfide gas.
6.0 L of gas in a piston at a pressure of 1.0 atm are compressed until the volume is 3.5 L. What is the new pressure inside the piston?
Answer:
1.7 atm
Explanation:
The formula you would want to use it P2=p1v1/v2
Plug in the numbers and solve
(1)(6)/3.5
Let me know if you need any other help!
Using Boyle's Law, we calculate the new pressure in the piston to be approximately 1.71 atm when the volume is decreased from 6.0 L to 3.5 L.
Explanation:The problem at hand relates to Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. In this instance, initial pressure (P1) is 1.0 atm, initial volume (V1) is 6.0 L, and the final volume (V2) is 3.5 L. The final pressure (P2) can be found by applying the formula from Boyle's Law: P1 x V1 = P2 x V2. Solving for P2, we find P2 = (P1 x V1) / V2 = (1.0 atm x 6.0 L) / 3.5 L, which simplifies to approximately 1.71 atm.
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A characteristic feature of any form of chromatography is the ________.a. calculation of an Rf value for the molecules separated. b. use of an inert carrier gas. c. use of a mobile and a stationary phase. d. use of molecules that are soluble in water.
Answer : The correct option is, (c) use of a mobile and a stationary phase.
Explanation :
Chromatography : It is a separation process or technique of a mixture in which a mixture is distributed between the two phases at different rates, one of which is stationary phase and another is mobile phase.
Mobile phase : The mixture is dissolved in a solution is known as mobile phase.
Stationary phase : It is an adsorbent medium and It is a solid, liquid or gel that remains immovable when a liquid or a gas moves over the surface of adsorbent. It remains stationary.
Hence, a characteristic feature of any form of chromatography is the use of a mobile and a stationary phase.