Answer:
The atmosphere refers to the gaseous envelope of earth, comprised of variable gases with definite proportions. The layers of the earth's atmosphere are as follows-
Troposphere- This layer starts from the ground and extends up to a height of about 10 km. Here, the temperature decreases with the increasing altitude. All the weather phenomenon takes place in this layer. Stratosphere- It starts from 10 km and extends up to a height of about 50 km. Here the temperature increases as the altitude increase. This is because of the presence of the ozone layer that receives the harmful UV radiation emitted from the sun. Mesosphere- This layer extends from a height of about 50 km to about 80 km above the earth's surface. Here, again the temperature decreases with the increasing altitude. Thermosphere- This layer starts from a height of about 80 km and extends up to about 500 km above the ground surface. In this region again the temperature increases with the increasing elevation. Exosphere- This layer ranges from about 500 km to 10,000 km above the earth's surface. Here, the temperature gradually increases with the increasing height.This variation in temperature occurs because of the certain reason. In the troposphere and the mesosphere, the temperature decreases with height because the pressure and height are inversely proportional to each other. The stratosphere experiences increasing temperature because of the presence of the ozone layer that is responsible for holding the greenhouse gases and the harmful UV radiation. The thermosphere and the exosphere experience high temperatures because of the receiving of the direct sunlight. Due to these above reason, there occurs this temperature change in a unique pattern.
The temperature changes within the layers of the atmosphere follow a pattern of alternating increases and decreases. This unique pattern is influenced by factors such as sun angle, insolation intensity, day length, and proximity to water bodies.
Explanation:The pattern of temperature changes within the layers of the atmosphere is characterized by alternating increases and decreases in temperature. This pattern can be observed in the five layers of the atmosphere, namely the troposphere, stratosphere, mesosphere, thermosphere, and exosphere. The temperature gradient varies with each layer, with the troposphere being the coldest and the thermosphere being the hottest. The unique pattern of temperature changes is caused by various factors such as sun angle, insolation intensity, day length, and proximity to water bodies.
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next few minutes?
A. Molecules in both the metal and the surrounding air will start moving at lower speeds.
B. Molecules in both the metal and the surrounding air will start moving at higher speeds.
C. The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
D. The air molecules that are surrounding the metal will speed up, and the molecules in the metal will s
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
The hot metal at 150 °C loses heat energy by conduction to the surrounding air molecukes and as such cools down, the cooler metal consists of lesser enery to power the movement of the molecules of the metal hence the metal molecules slows down in their movement as seen in the equation of heat and temperature
ΔH = m×C×ΔT where ΔH is the change in heat energy (heat loss of the metal, C is the heat capacity and ΔT is the temperature change
For the surrounding air that experiences increase in temperature the same process follows leading to increase in the kinetic energy of the air molecules and decrease in kinetic energy of the metal molecules as shown in the formula
K = [tex]\frac{3}{2}[/tex]×[tex]\frac{R}{N_{A} }[/tex]×T where K = Kinetic Energy, R = gas constant (8.314J/mol×K) and [tex]N_{A}[/tex] = Avogadros number (6.022×[tex]10^{23}[/tex] atoms/mol)
Answer:
d
Explanation:
Doc Inmaking is thinking about the density of liquid water (d = 0.99823 g/mL at 20 °C) compared to the density of ice (0.9168 g/mL at 0 °C). His favorite water bottle has a total volume of 300 mL. He fills it with exactly 288 mL of water at 20 °C. He tightens the lid and puts the bottle in the freezer. What mass of water did Doc placed in the bottle? (Three significant digits, unit of g.)
Answer:
The mass of water did Doc placed in the bottle is 288 grams.
Explanation:
Mass of water filled in water bottle = M
Volume of the water filled in water bottle = V = 288 mL
Density of the water at 20°C , d= 0.99823 g/ml
[tex]D=\frac{M}{V}[/tex]
[tex]M=D\times V=0.99823 g/ml\times 288 mL=287.49024 g\approx 288 g[/tex]
The mass of water did Doc placed in the bottle is 288 grams.
Final answer:
The mass of water at 20 °C with a density of 0.99823 g/mL that Doc placed in the bottle is 287 g when rounded to three significant digits.
Explanation:
The mass of water Doc placed in the bottle can be calculated using the density of liquid water at 20 °C, which is 0.99823 g/mL. Given he filled it with exactly 288 mL of water, the mass of the water is calculated by multiplying the volume by the density: The mass of water at 20 °C with a density of 0.99823 g/mL that Doc placed in the bottle is 287 g when rounded to three significant digits.
Mass = volume × density
Mass = 288 mL × 0.99823 g/mL
Mass = 287.63064 g
Since we are asked to provide the mass to three significant digits, the mass of water Doc placed in the bottle is 287 g.
The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-. In several experiment, the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed resulting in a straight line with a slope value of –1.10 x 104 K and a y-intercept of 33.5.
Answer and Explanation:
The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation
R = molar gas constant
K = A(e^(-Ea/RT))
Taking natural log of both sides
In K = In A - (Ea/RT)
In K = (-Ea/R)(1/T) + In A
Comparing this to the equation of a straight line; y = mx + c
y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A
a) From the question, m = (-Ea/R) = -1.10 × (10^4) K
(-Ea/R) = -1.10 × (10^4) = -11000
R = 8.314 J/K.mol
Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol
b) c = In A = 33.5
A = e^33.5 = (3.54 × (10^14))/s
c) K = A(e^(-Ea/RT))
A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol
K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s
QED!
This question pertains to the rate of a chemical reaction, which is first order with respect to (CH3)3CBr and zero order with respect to OH-. An Arrhenius plot indicating temperature and rate constant is used to find activation energy and frequency factor.
Explanation:The reaction you're describing is a typical chemical reaction. This type of reaction is first order with respect to (CH3)3CBr, which means the rate of the reaction depends on the concentration of this compound. OH-, on the other hand, is zero order, meaning its concentration doesn't affect the reaction's rate.
The plot you've mentioned is an Arrhenius plot, and it is used to determine the activation energy and frequency factor of a reaction from the slop and y-intercept respectively. Given the slope value of –1.10 x 10^4 K you mentioned, you can find the activation energy (Ea) from the formula Ea = -slope * R , where the slope is –1.10 x 10^4 K and R is the universal gas constant (8.3145 Joule/(mole*K)). Similarly, from the y-intercept value, you can find the frequency factor by the formula A=e^(y-intercept), where A is frequency factor and e is natural base.
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Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. What is the density of air at 22 °C and 760 torr? Assume ideal behavior.
Answer:
The density of air at 22 °C and 760 torr is 1.195 KG/m³
Explanation:
The solution to the above question is arrived t by considering the given variables and calculating the number of moles in a 1 m³ sample of air by plugging values into the universal gas equation from which the number of moles of the constituent gases can be calculated by Dalton's law of partial pressure, then their masses and lastly the density of air is calculated using the formula, Density = mass/volume
The given variables are
Percentage Nitrogen = 78.1% by volume
Percentage oxygen = 20.9% by volume
Percentage argon = 0.934% by volume
The molar mass of nitrogen = 14.006g/mol
The molar mass of oxygen = 15.999g/mol
The molar mass of argon = 39.948 g/mol hence Considering a unit volume of air of one cubic meter (1m^3) we have
0.781 m³ of nitrogen, 0.209 m³ of oxygen and 0.00934 m³ of argon
The number of moles in 1 m³ of gas at 22 °C and 760 torr is given by
PV = nRT or n = [tex]\frac{PV}{RT}[/tex] = where 760 torr = 101325Pa we have n = [tex]\frac{(101325)(0.001)}{(8.314)(295.15)}[/tex] = 0.00413 mols per liter or 41.29 moles/m³
thus we have number of moles of nitrogen = 42.129 × 78.1% = 32.25 moles and the mass of nitrogen = 32.25×28.02 = 903.6 g
number of moles of oxygen= 42.129 × 20.1% = 8.63 moles and the mass of nitrogen = 8.63×32 = 276.16 g
number of moles of argon= 42.129 × 0.934% = 0.386 moles and the mass of nitrogen = 0.386×40 = 15.43 g
Therefore, mass of one cubic meter of air (1 m³), has a mass of
903.6 g + 276.16 g + 15.43 g = 1195.2 g or 1.195 KG Hence the density of air
is given by Density = [tex]\frac{mass}{volume}[/tex] =[tex]\frac{1.195 KG}{1 m^{3} }[/tex] = 1.195 KG/m³
The density of air at 22 °C and 760 Torr is 1.19 g/L.
Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. We will calculate the average molar mass of the air (M) as a weighted average of the molar masses of its constituents.
[tex]M = 78.1\% \times M(N_2) + 20.9\% \times M(O_2) + 0.934\% \times M(Ar)\\\\M = 78.1\% \times 28.00g/mol + 20.9\% \times 32.00g/mol + 0.934\% \times 39.95 g/mol = 28.93 g/mol[/tex]
Then, we will convert 22 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 22\° C + 273.15 = 295 K[/tex]
Assuming ideal behavior, we can calculate the density (ρ) of the air at 295 K (T) and 760 Torr (P) using the following expression.
[tex]\rho = \frac{P \times M }{R \times T} = \frac{760 Torr \times 28.93g/mol }{(62.4mmHg.L/mol.K) \times 295K} = 1.19 g/L[/tex]
where,
R: ideal gas constantThe density of air at 22 °C and 760 Torr is 1.19 g/L.
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Which combination of formula, IUPAC name, and common name below is incorrect? Formula IUPAC Name Common Name (a) CHCl3 trichloromethane chloroform (b) CCl4 tetrachloromethane carbon tetrachloride (c) C6H5I iodobenzene phenyl iodide (d) CH3Cl chloromethane methyl chloride (e) CH2Cl2 dichloromethane methene chloride
Answer:
option (e), dichloromethane, methene chloride
Explanation:
(a) [tex]CHCl_3[/tex]
Common name: chloroform
IUPAC name: one carbon atom, therefore, root word is meth.
Its is saturated compounds, so add ane after root word.
Three chlorine atoms are present.
Therefore, IUPAC name: Trichloromethane
(b) [tex]CCl_4[/tex]
Common name: Carbon tetrachloride
IUPAC name: one carbon atom, therefore, root word is meth.
Its is saturated compounds, so add ane after root word.
Four chlorine atoms are present, chlorine atoms are named as prefixes.
Therefore, IUPAC name: tetrachloromethane
(c) [tex]C_6H_5I[/tex]
Common name: Phenyl iodide
IUPAC name:
The given compound is an aryl halides. Aryl haildes are named as haloarenes. The prefix halo is placed before aromatic hydrocarbon. Here, prefix is iodo and aromatic hydrocarbon is benzene.
Therefore, IUPAC name of the compound is iodobenzene.
(d) [tex]CH_3Cl[/tex]
Common name: Methyl chloride
IUPAC name: one carbon atom, therefore, root word is meth.
Its is saturated compounds, so add ane after root word.
One chlorine atom is present.
Therefore, IUPAC name: chloromethane
(e) [tex]CH_2Cl_2[/tex]
Common name: Methylene chloride
IUPAC name: one carbon atom, therefore, root word is meth.
Its is saturated compounds, so add ane after root word.
Two chlorine atoms are present, chlorine atoms are named as prefixes.
Therefore, IUPAC name: dichloromethane.
Therefore, the correct option is option (e), dichloromethane, methene chloride
Liquid 1 reacts with Liquid 2, producing a solid and a gas. Using this scenario, which supports the law of conse
mass?
mass of Liquid 1 + mass of solid = mass of Liquid 2 + mass of gas
mass of Liquid 1 - mass of solid = mass of Liquid 2-mass of gas
mass of Liquid 1 - mass of Liquid 2 = mass of solid + mass of gas
mass of Liquid 1 + mass of Liquid 2 = mass of solid + mass of gas
Answer:
The answer to your question is the fourth one.
Explanation:
From the description, we know that the reactants are two liquids and the products are a solid and a gas.
The first option is incorrect because it mentions that the reactants are one liquid and solid.
The second option is also incorrect for the same reason as the first one and one of the products must be a solid.
The third option is incorrect besides there are two liquids in the reactants and a solid and gas in the liquids there is a minus sign that is not possible.
The fourth option is the correct one.
The density of water at 3.98°C is 1.00000 g/mL. What is the mass in pounds of 16.743 L of water?
Answer:
The answer to your question is 36.9 pounds
Explanation:
Data
density = 1 g/ml
mass = ?
volume = 16.743 L
- To solve this problem use the formula of density.
density = mass / volume
- Solve for mass
Mass = density x volume
- Convert volume to ml
1000 ml --------------- 1 L
x ---------------- 16.743 L
x = (16.743 x 1000) / 1
x = 16743 ml
- Substitution
Mass = (1 g/ml)(16743 ml)
Simplification and result
Mass = 16743 g
- Convert mass to pounds
1 pound ------------------ 453.58 g
x ------------------ 16743 g
x = (16743 x 1) / 453.58
x = 36.9 pounds
The mass of 16.743 L of water in pounds can be calculated as approximately 36.904 pounds.
Explanation:To calculate the mass of water in pounds, we first determine the mass in grams using the given water volume and density.
Since the density of water is 1.00000 g/mL, and we know that 1 L is equal to 1000 mL, we multiply the volume in liters by the density in g/mL and by 1000 to get mass in grams.
This gives us 16.743 L * 1.00000 g/mL * 1000 = 16743 grams.
Converting grams to pounds, we know 1 pound is approximately 453.592 grams
So, the mass of water is then 16743 g / 453.592 g/lbs = 36.904 lbs. Therefore, the mass of 16.743 L of water is approximately 36.904 pounds.
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Suppose 0.410 kg of hexane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. Calculate the volume of carbon dioxide gas that is produced.Round your answer to 3 significant digits.
Answer:
The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane
Explanation:
We first write a balanced reaction for the complete combustion of hexane thus
The stoichiometry of the cumbustion of hexane in air is
2C6H14(g)+18O2(g)→12CO2(g)+14H2O(l) or
C6H14(g)+9O2(g)→6CO2(g)+7H2O(l)
From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide
Therefore we calculate the nuber of moles of hexane present in the sample thus
Mass hexane of sample = 0.41 kg
Molar nass of hexane = 86.18 g/mol
number of moles of hexane = (mass of hexane)/(molar mass of hexane) = (0.41×1000)/86.16 = 410/86.16 = 4.76 moles
As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces
4.76×6 moles of CO2 which is 28.55 moles of CO2
From the question we have the temperature and the pressure of the production of CO2 as
Temperature of reaction = 13° C converting to kelving gives= 13+273.15 = 286.15 K
and pressure = 1 atmosphere or 101325 Pa
13.0∘C=13.0∘C+273.15=286.15 K
The volume of the produced CO2 can be calculated using the combined ideal gas equation given by
P×V=n×R×T where
Here
P = Gas pressure (of CO2 )
V = Volume (of the CO2)
n = number of moles of gas (CO2) present
R = universal gas constant, equal to 0.0821 atm× L/(mol× K )
T = absolute temperature in Kelvin
Thus we have
1×V = 28.55×0.0821×286.15 or V = 670.76L
Rounding up the answer to 3 significant digits we have
670.76L ≅ 671L
671 litres of carbon dioxide is produced from 0.410 kg of hexane
At -18.6 °C, a common temperature for household freezers, what is the maximum mass of sucralose (C₁₂H₁₉Cl₃O₈) in grams you can add to 2.00 kg of pure water and still have the solution freeze?
Assume that sucralose is a molecular solid and does not ionize when it dissolves in water.
[tex]K_f[/tex] = 1.86 °C/m.
Answer : The maximum mass of sucralose is, 7952.8 grams.
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] for water = [tex]1.86^oC/m[/tex]
Mass of water (solvent) = 2.00 kg
Molar mass of sucralose = 397.64 g/mole
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of sucralose}}{\text{Molar mass of sucralose}\times \text{Mass of water in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = [tex]-18.6^oC[/tex]
[tex]\Delta T^o[/tex] = freezing point of water = [tex]0^oC[/tex]
i = Van't Hoff factor = 1 (for sucralose non-electrolyte)
[tex]K_f[/tex] = freezing point constant for water = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex](0-(-18.6)^oC)=1\times (1.86^oC/m)\times \frac{\text{Mass of sucralose}}{397.64g/mol\times 2.00kg}[/tex]
[tex]\text{Mass of sucralose}=7952.8g[/tex]
Therefore, the maximum mass of sucralose is, 7952.8 grams.
Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt MnSO 4 ( s ) . A 16.260 g sample of the hydrate was heated to 300 ∘ C . The resulting MnSO 4 ( s ) had a mass of 14.527 g . Calculate the value of n in MnSO 4 ⋅ n H 2 O ( s ) .
Answer:
Value of n in MnSO₄.nH₂O is one.
Explanation:
The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,
Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)
Mass of H₂O = 16.260 g - 14.527 g
Mass of H₂O = 1.733 g
moles of H₂O = (1.733 g) ÷ (18.015 g/mole)
moles of H₂O = 0.0962
For the moles of anhydrous salt:
moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄
moles of MnSO₄ = 14.5277 ÷ 151.001
moles of MnSO₄= 0.0962
Now for n:
n = moles of water ÷ moles of MnSO₄
n = 0.0962 ÷ 0.0962
n = 1
The above calculations show that one mole of H₂O is attached to the one formula unit of MnSO₄
Final answer:
To calculate the value of n in the hydrate formula MnSO₄ .nH₂O, the mass of water driven off by heating is found to be 1.733 g. This corresponds to 0.0962 mol of water. Since the ratio of water to MnSO₄is 1:1, n is determined to be 1.
Explanation:
To calculate the value of n in the hydrate formula MnSO₄ . H₂O, we need to find the number of moles of water lost upon heating. We subtract the mass of the anhydrous salt (14.527 g) from the original mass of the hydrate (16.260 g) to get the mass of water lost:
Mass of H₂O = 16.260 g - 14.527 g = 1.733 g
Next, we calculate the number of moles of water using its molar mass (18.015 g/mol):
Number of moles of H₂O = 1.733 g / 18.015 g/mol ≈ 0.0962 mol
To find the number of moles of MnSO₄ in the anhydrous sample, we need its molar mass, which is approximately 151.00 g/mol for MnSO₄. Using the mass of the anhydrous salt:
Number of moles of MnSO₄ = 14.527 g / 151.00 g/mol ≈ 0.0962 mol
This indicates that the mole ratio of H₂O to MnSO₄ is 1:1. Therefore, the value of n is 1, and the hydrate is MnSO₄.H₂O.
Characteristics of this mineral: hardness of 3, three directions of cleavage (cleaves into rhombs), partially transparent, effervesces with dilute HCl without being pulverized into a fine powder.
Answer:
Calcite
Explanation:
Calcite is a mineral formed by calcium carbonate (CaCO3), class 05 of the Strunz classification, the so-called carbonate and nitrate minerals.
All members of this group crystallize in the trigonal system, have a perfect rhombohedral cleavage and exhibit a strong double refraction in transparent rhombohedra.
It presents a variety of shapes and colors. It is characterized by its low hardness, 3 on the Mohs scale, and by its high reactivity even with weak acids.
The best property to identify calcite is the acid test, since this mineral produces effervescence with acids. The reason for this is the following reaction:
CaCO3 + 2H + → Ca2+ + H2O + CO2 (gas)
Select the statements that best describe the properties of an intramolecular sn2 reaction mechanism.
Intramolecular Sn2 reaction is a bimolecular, second-order, elementary reaction. It involves a single, concerted step in which a nucleophile attacks the substrate, leading to a transition state, and then to expulsion of a leaving group. The stereochemistry of the molecule is usually inverted at the reaction centre.
Explanation:To best describe the properties of an intramolecular Sn2 reaction mechanism, we can say that it's a bimolecular reaction, which means it involves two reactant species. In this case, the reaction mechanism involves a single, concerted step where a nucleophile attacks the substrate, leading to a transition state and finally the expulsion of a leaving group. This makes Sn2 a type of elementary reaction.
In these reactions, the rate is dependent on the concentration of both reactants, leading to a second-order rate law. Further, the rate-determining step (the slowest in the mechanism) for an Sn2 reaction is the single concerted step itself. One important aspect to remember about Sn2 mechanisms is the stereochemical alteration that takes place, typically resulting in inversion of configuration at the reaction centre.
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The intramolecular SN2 reaction mechanism is a bimolecular and concerted process characterized by inversion of configuration at the reaction center, second-order kinetics, and sensitivity to steric hindrance.
The intramolecular SN2 reaction mechanism is characterized by several distinct properties. Firstly, it is a bimolecular reaction, meaning the rate of the reaction depends on the concentration of two reactants: the nucleophile and the electrophile. Secondly, the reaction proceeds via a concerted process where bond-forming and bond-breaking occur simultaneously, leading to an inversion of configuration at the carbon center where the substitution takes place.
Lastly, the SN2 mechanism exhibits second-order kinetics, as the reaction rate depends on the concentration of both the nucleophile and the electrophile. It is important to note that SN2 reactions are sensitive to steric hindrance; bulky groups near the reactive site can inhibit the reaction by limiting the nucleophile's access to the electrophile.
es the molecule defective leading to sickle cell anemia. Predict whether the following hypothetical change would or would not have a major effect at position 6.
In hemoglobin, a single amino acid change at position 6 from Glu to Val has major consequences on hemoglobin structure that makes the molecule defective leading to sickle cell anemia. Predict whether the following hypothetical change would or would not have a major effect at position 6. Briefly explain (1-2 sentences). Glu to Leu Hint: Look at the structures of the R groups and consider their chemical properties
Answer:The structure of the haemoglobin, hence the RBC won't be same as normal.
Explanation:Both the leucine and glutamic acid are alpha amino acids which have an alpha carboxylic acid group and an alpha amino group. The variable in case of glutamic acid is propyl acid while the variable in case of leucine is isobutyl.
The glutamic acid is the normal amino acid of the 6th position of Beta chain of hemoglobin. Its an acid group, so can form bonds with another base inside the haemoglobin, or can form other hydrogen bonds. But the isobutyl group is an alkyl group. So it doesn't have that much effect in the recovering the structure, and sickle cell anemia prevails.
The interactions between water molecules and other non-water molecules through hydrogen bonding is known as ____________________________.
Answer: adhesion
Explanation:
Cohesion is the attraction between similar molecules. Example: Force of attraction between water molecules.
Thus hydrogen bond formed between the molecules of water due to the development of partial negative charge on oxygen and partial positive charge on hydrogen is cohesion.
Adhesion is the attraction between different molecules. Example: Force of attraction between HCl and water.
The hydrogen bond formed between H of HCl and O of water due to developments of partial positive and partial negative charge respectively is adhesion.
A compound's molecular formula must always be different than the compound's empirical formula. TRUE FALSE
Answer: False
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.
Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
Example: [tex]CH_4[/tex] has similar molecular formula and empirical formula as the elements are already present in simplest of the ratios.
[tex]C_2H_2[/tex] has molecular formula of [tex]C_2H_2[/tex] but [tex]CH[/tex] as the empirical formula.
Molecules can be described as
A) mixtures of two or more pure substances
B) mixtures of two or more elements that has a specific ratio between components
C) two or more atoms chemically joined together
D) heterogeneous mixtures
E) homogeneous mixtures
E) two or more atoms chemically joined together.
The following information should be considered:
A molecule refers to the smallest particle of a large compound that is created via the bonding of two or more atoms. The bonding should be between the atoms changes the physical and chemical properties of the particles.Learn more: https://brainly.com/question/2289757?referrer=searchResults
Answer:
two or more atoms chemically joined together.
Explanation:
An unknown element is found to contain isotopes with the following masses and natural abundances: 38.9637 amu (93.08%), 39.9640 amu (0.012%), and 40.9618 amu (6.91%). Using these data, identify the element. a. S b. K c. Cld. Ca e. Ar
Answer: The unknown element is potassium.
Explanation:
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex] .....(1)
For isotope 1:Mass of isotope 1 = 38.9637 amu
Percentage abundance of isotope 1 = 93.08 %
Fractional abundance of isotope 1 = 0.9308
For isotope 2:Mass of isotope 2 = 39.9640 amu
Percentage abundance of isotope 2 = 0.012 %
Fractional abundance of isotope 2 = 0.00012
For isotope 3:Mass of isotope 3 = 40.9618 amu
Percentage abundance of isotope 3 = 6.91 %
Fractional abundance of isotope 3 = 0.0691
Putting values in equation 1, we get:
[tex]\text{Average atomic mass of Z}=[(38.9637\times 0.9308)+(39.9640\times 0.00012)+(40.9618\times 0.0691)][/tex]
[tex]\text{Average atomic mass of Z}=38.85amu[/tex]
For the given options:
Option a: Average atomic mass of Sulfur = 32.065 amu
Option b: Average atomic mass of Potassium = 39.09 amu
Option c: Average atomic mass of Chlorine = 35.45 amu
Option d: Average atomic mass of Calcium = 40.078 amu
Option e: Average atomic mass of Argon = 39.94 amu
As, the average atomic mass of unknown element is near to the average atomic mass of potassium. So, the unknown element is potassium.
Hence, the unknown element is potassium.
The element with isotopes of masses 38.9637 amu, 39.9640 amu, and 40.9618 amu, and natural abundances of 93.08%, 0.012%, and 6.91% respectively, is Calcium (Ca).
Explanation:To identify the element based on its isotopes, we need to calculate the average atomic mass of the element. This can be done by multiplying the mass of each isotope by its natural abundance (expressed as a decimal), and then summing up these products. In this case, the element with isotopes of masses 38.9637 amu, 39.9640 amu, and 40.9618 amu, and natural abundances of 93.08%, 0.012%, and 6.91% respectively, is Calcium (Ca). The average atomic mass of calcium can be calculated as follows:
(38.9637 amu * 0.9308) + (39.9640 amu * 0.00012) + (40.9618 amu * 0.0691) = 40.08 amu
Therefore, the element is calcium (Ca).
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Determine the percent composition of CH2O.
Answer:
The given chemical compound has 2 atoms of hydrogen and one atom of oxygen for each atom of carbon. The mass of CH2O is 12 + 2*1 + 16 = 30. The molecular weight of the compound is 180.18 which is approximately 180. This gives the molecular formula of the chemical compound as C6H12O6.
Explanation:
To calculate the percent composition of CH2O, determine the molar mass of each element and the total molar mass of the compound. The percent composition is approximately 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.
Explanation:To calculate the percent composition of CH2O, we will need to determine the molar mass of each element in the compound and the total molar mass of the compound. The molar masses from the periodic table are approximately 12.01 g/mol for Carbon (C), 1.01 g/mol for Hydrogen (H), and 16.00 g/mol for Oxygen (O).
First, let's calculate the total molar mass of CH2O: (1 × 12.01) + (2 × 1.01) + (1 × 16.00) = 12.01 + 2.02 + 16.00 = 30.03 g/mol.
Now, let's find the percent composition for each element:
Carbon: (12.01 g/mol ÷ 30.03 g/mol) × 100% = 40.0%Hydrogen: (2.02 g/mol ÷ 30.03 g/mol) × 100% = 6.7%Oxygen: (16.00 g/mol ÷ 30.03 g/mol) × 100% = 53.3%The percent composition of CH2O is therefore approximately 40.0% Carbon, 6.7% Hydrogen, and 53.3% Oxygen.
Mathematically combine the three given reactions so that they sum to give a balanced chemical equation describing the production of propane C3H8(g), from its elements, C(s.graphite) and H2lg). Show your intermediate steps. For example, if you must reverse reaction (c), enter the reverse in the appropriate answer box; if you multiply reaction (a) by 2, enter the updated equation in the corresponding answer box. Make sure to enter the overall balanced equation at the bottom (a) C3Hg(g)+502C2+H2O0) (b) C(s)+02(g)> CO2(g) (c) H2(«)+ 7o(Hod)
Answer and Explanation
The final reaction is the production of propane from Carbon and Hydrogen.
3C (s) + 4H2 (g) ---> C3H8 (g)
So, reverse eq. A,
3CO2 (g) + 4H20 (l) ---> C3H8 (g) + 502 (g)
Add 3 × eq. B,
3C (s) + 3O2 (g) ----> 3CO2 (g)
Add 4 × eq. C,
4H2 (g) + 2O2 (g) ---> 4H20 (l)
Writing them together,
3CO2 (g) + 4H20 (l) ---> C3H8 (g) + 502 (g)
+ 3C (s) + 3O2 (g) ----> 3CO2 (g)
+ 4H2 (g) + 2O2 (g) ---> 4H20 (l)
------------------------------------------------------------
3CO2 (g) + 4H20 (l) + 3C (s) + 3O2 (g) + 4H2 (g) + 2O2 (g) ---> C3H8 (g) + 502 (g) + 3CO2 (g) + 4H20 (l)
The compounds that exist on both sides cancel out and we're left with
3C (s) + 4H2 (g) ---> C3H8 (g)
So, mathematically, the final reaction can be written as:
(-eq. A + 3(eq. B) + 4(eq. C))
-A+3B+4C = 3C (s) + 4H2 (g) ---> C3H8 (g)
QED!
[tex]\[{3\text{C(s, graphite)} + 4\text{H2(g)} \rightarrow \text{C3H8(g)} + 5\text{O2(g)}} \][/tex]
This balanced equation shows the production of propane (C3H8) from its elements carbon and hydrogen gas.
To combine the given reactions to form the balanced chemical equation describing the production of propane (C3H8) from its elements carbon (C, as graphite) and hydrogen (H2), let's follow these steps:
Given reactions:
(a) [tex]\( \text{C3H8(g)} + 5\text{O2(g)} \rightarrow 3\text{CO2(g)} + 4\text{H2O(g)} \)[/tex]
(b) [tex]\( \text{C(s, graphite)} + \text{O2(g)} \rightarrow \text{CO2(g)} \)[/tex]
(c) [tex]\( \text{H2(g)} + \frac{1}{2}\text{O2(g)} \rightarrow \text{H2O(g)} \)[/tex]
We need to manipulate these reactions to combine them into one overall balanced equation that shows the formation of propane (C3H8) from carbon and hydrogen.
Step-by-Step Combination:
1. Reverse Reaction (a):
[tex]\( 3\text{CO2(g)} + 4\text{H2O(g)} \rightarrow \text{C3H8(g)} + 5\text{O2(g)} \)[/tex]
This is the reverse of reaction (a), which is necessary to show the formation of propane.
2. Multiply Reaction (b) by 3:
[tex]\( 3\text{C(s, graphite)} + 3\text{O2(g)} \rightarrow 3\text{CO2(g)} \)[/tex]
Multiply reaction (b) by 3 to balance the carbon atoms with the propane formation reaction.
3. Multiply Reaction (c) by 4:
[tex]\( 4\text{H2(g)} + 2\text{O2(g)} \rightarrow 4\text{H2O(g)} \)[/tex]
Multiply reaction (c) by 4 to balance the hydrogen atoms with the propane formation reaction.
4. Combine the Reactions:
Now, add the balanced reactions (reverse of (a), multiplied (b), and multiplied (c)) to get the overall balanced equation for the formation of propane:
[tex]\( 3\text{C(s, graphite)} + 3\text{O2(g)} + 4\text{H2(g)} + 2\text{O2(g)} \rightarrow 3\text{CO2(g)} + 4\text{H2O(g)} + 5\text{O2(g)} \)[/tex]
5. Simplify the Equation:
Combine like terms (oxygen on both sides):
[tex]\( 3\text{C(s, graphite)} + 3\text{O2(g)} + 4\text{H2(g)} \rightarrow 3\text{CO2(g)} + 4\text{H2O(g)} + 5\text{O2(g)} \)[/tex]
A chemist began with 61.5 grams of naclo3.After collecting and drying the product, 30 grams of nacl was obtained. what was the theoretical yield of nacl
Answer:
33.78 g
Explanation:
NaClO3 decomposes to NaCl and O2 by this reaction:
2NaClO3 --> 2NaCl + 3O2
Let's determines the mole of chlorate we used (mass / molar mass)
61.5 g / 106.45 g/mol = 0.578 moles.
Ratio is 2:2, so x amount of chlorate will produce x amount of chloride. In conclussion we made 0.578 moles of NaCl from 0.578 moles of chlorate. Let's convert the moles to mass:
0.578 mol . 58.45g/1mol = 33.78 g
That is the theoretical yield of NaCl.
Over time, Hinduism has become/is becoming Select one: a. less homogeneous. b. more localized. c. more homogeneous. d. more racialized.
Answer:
d. Localized.
Explanation:
You posted this question in the chemistry section. The whole exercise requires you to understand the terms in a chemical context. So let's look at the three terms: homogeneous, localized, and radicalized.
Homogeneous
A chemical substance is said to be homogeneous if the chemical composition is the same throughout. It means the substance must have the same states. For example, a mixture of undiluted orange juice forms a homogeneous solution once it is diluted in water. It means that there is perfect mixing to have a uniform liquid state.
Localized
The terminology is often applied to atoms or chemical structures, particularly metals and acids. For example, atoms have localized electrons in the orbitals. It means that the electron orbits in the region for 95 % of its time. Localized means belonging to one region. This applies to the religion.
Radical
The term applies to highly reactive atomic species, normally called ions. These elements seek electrons and are highly "reactive." The ions are called radicals in this sense.
On a small farm, the weights of eggs that young hens lay are normally distributed with a mean weight of 51.3 grams and a standard deviation of 4.8 grams. Using the 68-95-99.7 rule, about what percent of eggs weigh between 46.5g and 65.7g.
Final answer:
Using the 68-95-99.7 rule, the percentage of eggs weighing between 46.5 grams and 65.7 grams, given a mean weight of 51.3 grams and a standard deviation of 4.8 grams, is estimated to be between 95% and 99%.
Explanation:
The question asks us to calculate the percentage of eggs that weigh between 46.5 grams and 65.7 grams, given that the weights are normally distributed with a mean of 51.3 grams and a standard deviation of 4.8 grams. Utilizing the 68-95-99.7 rule (also known as the Empirical Rule), we can determine percentages for different ranges from the mean in a normal distribution.
Firstly, to find the specific range that includes 46.5g to 65.7g from our mean of 51.3g, we calculate the number of standard deviations each value is from the mean. However, without doing the math, we see that 46.5g is less than one standard deviation away (since one standard deviation is 4.8g), and 65.7g is significantly more than two but less than three standard deviations away.
According to the 68-95-99.7 rule, 68% of data falls within one standard deviation, 95% within two, and 99.7% within three. Thus, intuitively, without precise calculation, we can say that the percentage of eggs weighing between 46.5g and 65.7g would be a bit less than 99.7%, as the upper limit is not yet reaching three standard deviations from the mean but is beyond the two-standard deviation mark that covers 95% of the distribution. Thus, it's reasonable to conclude that approximately 95-99% of eggs will fall within this weight range.
Some prokaryotes, such as the blue-green ____________ , are photosynthetic and contain ____________ where chlorophyll and other pigments absorb energy from the sun to produce carbohydrates via photosynthesis.
Blue-green algae are prokaryotes that carry out photosynthesis in chloroplasts, converting light energy into carbohydrates by using chlorophyll as a pigment for energy absorption.
Explanation:Some prokaryotes, such as blue-green algae, are photosynthetic and contain chloroplasts where chlorophyll and other pigments absorb energy from the sun to produce carbohydrates via photosynthesis. The algae, specifically, are among the groups of prokaryotes that are capable of photosynthesis, much like plants. The chloroplasts in these cells function as the site where light energy captured by chlorophyll is converted into chemical energy, which is then used to create carbohydrates from carbon dioxide and water.
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A railroad diesel engine weighs four times as much as a freight car. The diesel engine coasts at 5 km/h into a freight car that is initially at rest. Use the conservation of momentum to show that after they couple together, the engine + car coast at 4 km/h.
Please show work!!!! step by step
Explanation:
Conservation of momentum :
[tex]m_1u_1+m_2u_2=m_1v_1+m_1v_2[/tex]
Where :
[tex]m_1, m_2[/tex] = masses of object collided
[tex]u_1,u_2[/tex] = initial velocity before collision
[tex]v_1,v_2[/tex] = final velocity after collision
We have :
Mass of an engine = [tex]m_1=4M[/tex]
Mass of an car= [tex]m_2=M[/tex]
Initial velocity of railroad engine [tex]m_1=u_1=5 km/h[/tex]
Initial velocity of car [tex]m_2=u_2=0 km/h[/tex] (rest)
Final velocity of railroad engine [tex]m_1=v_1=v[/tex] (same direction )
Final velocity of car [tex]m_2=v_2=v[/tex] (same direction)
[tex]4M\times 5km/h+M(0 km/h)=4Mv+Mv[/tex]
[tex]4\times 5 km/h=5M[/tex]
v = 4 km/h
The speed of the engine and car after they coupled together is 4 km/h.
Using the principle of conservation of momentum, it's shown that the diesel engine and freight car will coast together at 4 km/h after coupling. This is calculated by equating the initial and final momentums of the system.
Explanation:The conservation of momentum can be used to solve this problem. The principle states that the total linear momentum of a closed system remains constant, regardless of any internal changes. Here, the system consists of the diesel engine and the freight car.
Let's denote the weight of the freight car as 'm'. Given that the diesel engine weighs four times as much as a freight car, the weight of the engine would be '4m'.
If the diesel engine is coasting at 5 km/hr, the initial momentum of the system is the momentum of the engine, because the freight car is at rest. Therefore, the initial momentum (Pi) is the weight of the engine times its velocity, which is 4m*5 km/hr = 20m km/hr.
After they couple together, there's no external force, so the total momentum should remain the same (the conservation of momentum principle). Let's denote the final velocity of the engine + car (now moving together) as 'v'. The final momentum (Pf) = (m + 4m) * v = 20m km/hr.
Therefore, we can establish the equation: Pi = Pf, meaning 20m km/hr = 5m * v. Solving for v, we find that v = 4 km/h. Therefore, the engine + car coast together at 4 km/h after coupling, demonstrating the conservation of momentum.
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Classify these substances? More than one answer may apply in each case.
N 2 solution heterogeneous mixture homogeneous mixture element compound pure substance
O 2 pure substance homogeneous mixture solution heterogeneous mixture element compound
N 2 O compound heterogeneous mixture element homogeneous mixture solution pure substance Air
(mostly N 2 and O 2 )
homogeneous mixture heterogeneous mixture solution pure substance element compound
Answer:
N2 element, pure substance
O2 element, pure substance
N2O Compound, pure substance
Air (mostly N2 and O2 ) homogeneous mixture
Explanation:
N2, Nitrogen is known as the chemical element that is characterized by having atomic number 7 and that is symbolized by the letter N, in its molecular version, it is recognized as N2.
O2, Oxygen is the chemical element of atomic number 8, this molecular form is composed of two atoms of this element.
A chemical element is a type of matter, consisting of atoms of the same class.
N2O, Nitrous oxide is formed by the union of two molecules of nitrogen and one of oxygen, which is considered a chemical compound since it is a substance formed by the chemical combination of two different elements of the periodic table.
A pure substance is one that cannot change state or divide into other substances, except for a chemical reaction.
Air (mostly N2 and O2 ), it is a homogeneous mixture of gases that constitutes the earth's atmosphere. A homogeneous mixture is a type of mixture in which its components are not distinguished and in which the composition is uniform and each part of the solution has the same properties.
Substances can be classified based on their composition and uniformity: N2 and O2 are elements and pure substances, N2O is a compound and a pure substance, and air is a homogeneous mixture or solution.
Explanation:When classifying substances, we take into account their composition and uniformity. Here are the classifications for the mentioned substances:
N2 (Nitrogen) is an element and a pure substance since it is composed of only one type of atom.
O2 (Oxygen) is also an element and a pure substance, with two oxygen atoms bonded together.
N2O (Nitrous Oxide) is a compound as it is made up of two different elements, nitrogen and oxygen, in a fixed ratio and a pure substance due to its uniform composition.
Air, which is mostly made up of Nitrogen (N2) and Oxygen (O2), is a homogeneous mixture or solution because the composition is uniform throughout, and it is a mixture of multiple gases.
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One isotope of carbon (C) has exactly the same mass number and atomic mass since it was used as the definition of the atomic mass unit (amu). Which isotope is it and what is its atomic mass?
Answer:The isotope is Carbon-12 and its atomic mass is 12.
Explanation:
Mass number is the total number of protons and neutrons in a nucleus.
Atomic number is the number of protons in the nucleus of an atom.
An isotope of a chemical element is an atom that has a different mass number but the same atomic number as the element. The difference in mass number is from the number of neutrons (that is, a greater or lesser atomic mass) than the standard for that element.
Carbon-12 is an isotope of carbon it has 6 neutrons and 6 protons, giving it a mass number of 12 and atomic number of 6. Carbon-12 is a stable isotope of carbon, it has the same mass number and atomic number as carbon.
The isotope of carbon that has exactly the same mass and atomic number as used in the definition of the atomic mass unit is;
Isotope Carbon-12.
In chemistry, we know that;
Mass number is defined as the sum of the protons and the neutrons present in the nucleus of an atom.Meanwhile, Atomic number is defined the number of protons in the nucleus or number of electrons around the nucleus of an atom.
While an isotope is defined as an atom of an element that has the same atomic number but different mass number.
The isotope Carbon-12 is an isotope of carbon that has 6 protons and 6 neutrons.This means from the definition of mass number; Carbon-12 will have a mass number = 6 + 6 = 12Number of protons = number of electrons. Thus, number of electrons = 6 and therefore, atomic number = 6. From periodic table, the element Carbon has the same mass number and atomic number as its' isotope carbon-12.
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During an experiment, 575 mL of neon gas at 101 kPa were compressed in a cylinder to a volume of 144 mL.
What was the new pressure of the gas, if the temperature remained constant?
Answer:
403.3 kPa is the new pressure
Explanation:
This problem is solved by this formula:
P₁ . V₁ = P₂ . V₂
101kPa . 575 mL = P₂ . 144mL
(101kPa . 575 mL) / 144 mL = P₂
403.3 kPa = P₂
Final answer:
Using Boyle's Law, the new pressure of neon gas compressed from 575 mL at 101 kPa to 144 mL, with temperature held constant, is found to be 403.125 kPa.
Explanation:
The student asked for the new pressure of neon gas that was compressed from 575 mL at 101 kPa to a volume of 144 mL, given that the temperature remained constant. This type of problem involves ideal gas behavior and can be solved using Boyle's Law, which states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure (P₁V₁ = P₂V₂).
Starting with the initial conditions:
Initial volume (V₁) = 575 mL
Initial pressure (P₁) = 101 kPa
And the final condition:
Final volume (V₂) = 144 mL
Since temperature remains constant, we can calculate the final pressure (P₂) using the formula:
P₂ = (P₁* V₁) /V₂
Substitute the given values into the equation:
P₂ = (101 kPa* 575 mL) \/ 144 mL = 403.125 kPa
Hence, the new pressure of the neon gas after compression is 403.125 kPa.
In aerobic cellular respiration, if four molecules of pyruvic acid enter steps two, the formation of acetyl CoA and three, the Krebs cycle, how many molecules of ATP, NADH, and FADH2 will be formed?
Answer:
The aerobic cellular respiration of the glucose where glucose is converted to energy via four steps as follows
1. Glycolysis (glucose break down to pyruvic acid)
2. Link reaction
3. Krebs cycle
4. Electron transport chain, or ETC
The four pyruvic acid produces Four ATP, twenty NADH, and four [tex]FADH_{2}[/tex] molecules
Explanation:
When four pyruvic acid enters step two of the aerobic cellular respiration, they are converted by Oxidative decarboxylation into acetyl-CoA, four molecules of NADH and four molecule of CO2 are formed. This process is otherwise called the link reaction or transition step, because it connects or links the Krebs cycle and glycolysis.
From the chemical reactions involved in cellular respiration of one glucose molecule, from two pyruvic acid molecules we have 2 ATP molecules, 10 NADH molecules, and 2 FADH2 molecules
Hence from four pyruvic acid molecules we have that the acetyl-CoA produced from the four pyruvic acid enters the the Krebs cycle and forms four ATP molecules, twenty NADH molecules, and four [tex]FADH_{2}[/tex] molecules.
In aerobic cellular respiration, four molecules of pyruvic acid will generate a total of four molecules of ATP, sixteen molecules of NADH (four from the conversion to Acetyl CoA, and twelve from the Krebs cycle) and four molecules of FADH2
Explanation:In aerobic cellular respiration, pyruvic acid is converted into acetyl CoA. One molecule of pyruvic acid generates one molecule of NADH during this conversion so four molecules of pyruvic acid will yield four molecules of NADH. Acetyl CoA then enters the Krebs cycle, for each molecule of Acetyl CoA that goes through the Krebs cycle, three molecules of NADH, one molecule of FADH2, and one molecule of ATP is formed. Therefore, the four molecules of pyruvic acid would end up generating four molecules of ATP, twelve molecules of NADH and four molecules of FADH2 during the Krebs cycle (not including the NADH generated during the conversion to Acetyl CoA).
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Ethanol contains the elements carbon, hydrogen, and oxygen. When ethanol burns, it chemically reacts with oxygen gas. C2H6O + O2 ethanol oxygen gas What elements will be present in the substances that are created when ethanol burns?
Answer:
The elements that will be present after the burning of Ethanol are;
(i) Carbon
(ii) Oxygen
(iii) Hydrogen
Explanation:
The balanced chemical equation for the burning of ethanol is as follow;
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
It can be observed in given balanced chemical equation that there three elements involved in this entire reaction. The elements that present in the reactant side are also found on the product side. It means that the elements have just rearranged going from reactant to product.
This means that this reaction is obeying Law of Conservation of Mass which states that mass can neither be created nor destroyed but it can be changed from one form to another hence, to keep the mass on both sides of the reaction balanced the same elements should be present on the product side too.
Answer: carbon, hydrogen, and oxygen only
Explanation: When a chemical reaction occurs, the atoms in the original set of substances are rearranged to form a new set of substances. The number of atoms of each element does not change.
The elements carbon, hydrogen, and oxygen are the only elements present in the original substances, so the only elements in the final substances after the reaction will be carbon, hydrogen, and oxygen, as well. The number of atoms of each element must be the same before and after the reaction.
Describe a sigma bond.a. orbital overlapping with the side of a p orbitalb. overlap of two d orbitalsend to end overlap of p orbitalc. sp orbital overlapping with an f orbitald. side by side overlap of d orbitals
A sigma bond b. overlap of two d orbitals end-to-end overlap of p orbital.
What is a sigma bond?A sigma bond is a strong bond that is made up of overlapping orbitals. In fact, these bonds are the strongest known bonds in chemical reactions. The overlapping orbitals are in the form of covalent bonding and this gives us the idea that sigma bonds are strong in nature.
In addition, sigma bonds are mostly common with diatomic elements and compounds. There are three orbitals where the sigma bond can be found and they are the p-p, s-p, and the s-s orbitals. These are known for forming symmetry groups.
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