White-tailed deer are mammals. Mammals have a reproductive strategy
that is not found in all animals.
Which two statements provide a cause and effect of the reproductive
strategy of white-tailed deer that helps them survive?
A. It requires two parents.
B. It requires only one parent.
C. It causes very fast growth and development.
D. It produces genetic variation within a population.
E. It results in individuals being identical to one parent.
Answers
This lack of genetic variation can lead to a decrease in the population’s ability to survive and adapt to changing environments, resulting in a population that is more at risk of extinction.
Which two statements provide a cause and effect of the reproductive strategy of white-tailed deer that helps them survive?A. It requires two parents. - CauseD. It produces genetic variation within a population. - EffectThe reproductive strategy of white-tailed deer helps them survive because it requires two parents.This helps ensure that the young deer receive the care and protection of both parents, which increases their chance of survival.Additionally, the two-parent strategy of the deer produces genetic variation within a population.This variation allows for natural selection to take place and for the population to adapt to changing environments. As a result, members of the population are more likely to survive and reproduce, creating a more resilient population.In contrast, a one-parent strategy, such as parthenogenesis, produces individuals that are identical to one parent.This lack of genetic variation can lead to a decrease in the population’s ability to survive and adapt to changing environments, resulting in a population that is more at risk of extinction.To learn more about The reproductive strategy refer to:
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Related Questions
One mole of tungsten (6 × 1023 atoms) has a mass of 184 grams, and its density is 19. 3 grams per cubic centimeter, so the center-to-center distance between atoms is 2. 51 × 10-10 m. You have a long thin bar of tungsten, 2. 6 m long, with a square cross section, 0. 07 cm on a side. You hang the rod vertically and attach a 98 kg mass to the bottom, and you observe that the bar becomes 1. 42 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in tungsten. 1) What is the spring stiffness of the entire wire, considered as a single macroscopic (large scale), very stiff spring?
Answers
The wire is 91826.77 N/m stiff, One layer of a cross section contains 1.0158e13 atoms, Number of bonds per unit length = 9.96e9, rigidity = 9.035 e-9 N/m
a) The force acting on the bar is given by (F) = mg = 119 * 9.8 = 1166.2 N.
The wire's extension (L) is 1.27 cm, or 0.0127.
Therefore, stiffness of wire = F/L = 1166.2 / 0.0127.
The wire is 91826.77 N/m stiff.
b) The ratio of the number of atoms in one cross-sectional layer to 8e-4 (2.51e-10)2
One layer of a cross section contains 1.0158e13 atoms.
c) The length's bond count is 2.5 / 2.51e-10.
Number of bonds per unit length = 9.96e9
d) Bond stiffness equals force acting on the bond / bond strain
bond-applied force = 1.152e-20 N
strain is equal to (0.0127/2.5)*2.51e-10 = 1.275e-12 m.
Stiffness is calculated as 1.152e-20 / 1.275e-12.
rigidity = 9.035 e-9 N/m
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33. A radioactive sample has an initial activity of 2.00 x 10^6cpm (counts per minute), and after 4.0 days, its activity is 9.0 x 10^5 cpm. What is its activity after 26 days?
A) 9.3 cpm
B) 1.1 x 10^4 cpm
C) 1.8 x 10^2 cpm
D) 2.8 x 10^-9 cpm
E) none of these
Please Show Work
Answers
The activity after 26 days can be calculated as 1.1 x 10^4 cpm.
What is radioactive?Radioactive is a type of energy that is emitted from certain elements when they undergo radioactive decay. Radioactive elements, such as uranium and radium, contain unstable atoms that are in a constant state of decay. When these atoms break apart, they release energy in the form of radiation.
The decay constant (k) of the sample can be calculated as follows:
k = ln(2.00 x 10^6/9.0 x 10^5) / 4.0
k = 0.073
Therefore, the activity after 26 days can be calculated as follows:
A = 2.00 x 10^6 e^(-0.073 x 26)
A = 1.1 x 10^4 cpm
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How does sunscreen work?(1 point) Responses .
It helps prevent the skin from absorbing UV waves.
It helps UV waves be able to pass through the skin
It helps UV waves be able to pass through the air.
It helps prevent the skin from reflecting UV waves.
Answers
Inorganic chemicals in sunscreen can reflect or scatter the light away from the skin, and organic (carbon-based) ones can absorb UV rays
Answer:
It helps prevent the skin from absorbing UV waves.
Explanation:
which is the correct molar mass for the compound cabr2?
Answers
199.86 0r 200g/mol molar mass for the compound cabr2
The orbitals or shells that enclose an atom's nucleus are home to electrons. In the nucleus, protons and neutrons are present. In the atom's nucleus, they coalesce. An atom's nucleus is surrounded by electrons in distinct orbits. The molar mass is calculated by dividing the mass (g) of a certain chemical element or chemical compound by the amount of the substance (mol). The standard atomic masses (in g/mol) of the component atoms can be added to determine the molar mass of a compound.
Thus, the molar mass of CaBr2 can be calculated as shown below.
Molar mass=(1×40.078gmol+(2×79.904gmol)
=40.078gmol+159.808gmol
=199.886gmol
Hence, the molar mass of CaBr2 is 199.886gmol.
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A sample of hydrogen has a volume of 12 L under a pressure of 3 atm. What will the pressure of this gas be if the volume were decreased to 6 L?
Answers
The pressure of the gas if the volume were decreased to 6L is 6 atm.
How to calculate pressure?Boyle's law of ideal gases states that the pressure of an ideal gas is inversely proportional to its volume at constant temperature.
The Boyle's law equation is given as follows;
P₁V₁ = P₂V₂
Where;
P₁ = initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
3 × 12 = 6 × P
36 = 6P
P = 6atm
Therefore, the pressure of the hydrogen gas is 6 atm.
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water is added to a 8.23 g sample of tacl5. the only products are 5.71g of a solid containing only tantalum, chlorine and oxygen and 3.35 g of a gas which is 97.2% chlorine and the remainder is hydrogen. (a) determine the empirical formula of the gas. (b) what fraction of the chlorine of the original compound is in the solid? (c) determine the empirical formula for the solid produced. (d) write a balanced equation for the reaction between tantalum pentachloride and water
Answers
The empirical formula is the simplest formula for a compound which is defined as the ratio of subscripts of smallest possible whole number of the elements present in the formula. It is also known as the simplest formula.
write a balanced equation for the reaction between tantalum pentachloride and water?
Tantalum Pentachloride is used as the chlorinating agent of the organic compound, chemical intermediates, and preparation as tantalum.TaCl5 is used in the preparation of catalyst for the polycyclotrimerizations of alkenediynes, chloro-aryloxide compounds and for the plasma-enhanced atomic layer deposition of tantalum nitride films. This product is involved in the preparation of tantalum(V) oxychloride.Tantalum oxide (Ta2O5) is one of the most important transition metal oxides because of its extraordinary physical and chemical properties, including high dielectric and refractive coefficients and excellent photoelectric performance.To learn more about chlorine refers to:
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A 55-kg woman has 7. 5 × 10−3 mol of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?
Answers
The number of hemoglobin molecules can be calculated by multiplying the number of moles of hemoglobin by Avogadro's number (N_A), which is the number of entities (such as atoms or molecules) in one mole of a substance.
So, the number of hemoglobin molecules in the woman's blood is:
7.5 × 10^-3 mol * 6.02 x 10^23 molecules/mol = 45.15 x 10^20 molecules
To convert the number of moles of hemoglobin to grams, we can multiply the number of moles by the molar mass of hemoglobin:
7.5 × 10^-3 mol * 64,456 g/mol = 483.17 g
So, the woman has approximately 483.17 grams of hemoglobin in her blood.
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a 5.250 mg sample of nicotine, composed entirely of c, h, and n, is combusted, producing 14.242 mg of co2 and 4.083 mg of h2o. what is the empirical formula of nicotine?
Answers
The empirical formula of nicotine is C0.96H1N1
The empirical formula of a compound is the simplest whole-number ratio of the atoms in the compound.
To find the empirical formula of nicotine, you need to determine the number of moles of each element present in the sample and then divide by the lowest number of moles.
First, you need to convert the mass of nicotine, CO2, and H2O to moles. To do this, you can use the molar mass of each element:
C = 12.01 g/mol
H = 1.01 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
5.250 mg of nicotine is equal to:
[tex]\frac{ 5.250 mg}{(12.01+1.01+14.01)g/mol}[/tex]) = 0.0027 moles
14.242 mg of CO2 is equal to:
[tex]\frac{14.242mg}{(12.01+16.00*2)g/mol}[/tex] = 0.0024 moles
4.083 mg of H2O is equal to:
[tex]\frac{4.083mg}{(1.01*2+16.00)g/mol}[/tex] = 0.0013 moles
Now you need to find the number of moles of each element present in the sample:
C = 0.0027 moles × 1
H = 0.0013 moles × 2
N = 0.0027 moles × 1
And finally, you can divide these values by the lowest number of moles to get the simplest whole-number ratio:
C = [tex]\frac{ 0.0027 moles}{0.0027 moles}[/tex] = 1
H = [tex]\frac{0.0026 moles}{ 0.0027 moles}[/tex] = 0.96
N = [tex]\frac{ 0.0027 moles}{0.0027 moles}[/tex] = 1
The empirical formula of nicotine is C0.96H1N1
It's important to note that the empirical formula represents the simplest ratio of atoms in a molecule but doesn't necessarily represent the actual molecular formula which could have more atoms or a different arrangement of atoms.
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what would happen to the percent yield of product if a student had not heated the test tube for as long as the directions instructed? explain.
Answers
The percent yield of product if a student had not heated the test tube for as long as the directions instructed is the result of the product will not match even the percent yield of the product being tested can fail.
A test tube is a tool used to carry out biochemical tests and grow microbes. The media that can be put into the test tube are solid media and liquid media and for the test tube lid you can use cotton, aluminum foil, metal caps, or plastic caps.
How to use a test tube can be heated into a beaker that has been filled with water and heated with a spirit burner for the heating process. Heating the test tube is important for laboratory tests such as purification of solvents by distillation and chemical reactions. One function of heating the solution is to increase the solubility of the solute in it. If a student does not heat the test tube according to the instructions given, the results of the product being tested will not be maximized. Even trials regarding the product fail with a high percentage.
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elect all the statements that correctly describe resonance structures.multiple select question.resonance structures differ only in the arrangement of electrons, not the atoms.an individual resonance structure does not accurately represent the structure of the species.resonance structures are isomers of the same species.resonance forms rapidly interconvert and the species could have any one of these structures at any time.individual resonance forms are not real.
Answers
All the statements given above are correct.
Atoms within resonance structures are the same. The species' structure is not accurately reflected by a single resonance structure.
What are resonance structures?A representation of a molecule or an ion called a resonance structure depicts several electron combinations that are not conceivable in a single, static structure.
Resonance hybrid refers to the actual structure of the molecule or ion, which is an average of the potential resonance structures.
Resonance structures are not different in terms of atoms, but rather in the arrangement of electrons.
Resonance structures vary, and no single resonance structure can fully capture the structure of an entire species.
The species that make up resonance structures have isomers.
The species could have any of these structures at any time since resonance forms quickly interconvert.
It is not real to have individual resonance shapes.
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What do you notice about all of the ionic charges for group 12 group 2 group 15 group 16 group 17
Answers
Answer: The elements in Groups 2,15,16 and 17 of the Periodic Table are called alkali metals. They form ionic compounds when they react with non-metals. Their ions have a single positive charge. For example, sodium forms sodium ions, Na+
Explanation:
hope you understood
What experimental evidence led Rutherford to develop his atomic model? What was the reasoning that led Rutherford to develop this model?
Answers
The experimental evidence that led Rutherford to develop his atomic model is; from a gold foil, alpha particles were shown to disperse backward.
Alpha particles were seen to disperse in the opposite direction after being emitted from a gold foil in the now-famous experiment. In his explanation, which he published in May 1911, Rutherford proposed that the dispersion was brought about by a rigid and compact core at the center of the atom known as the nucleus.
The reasoning that went into developing the atomic model was that the only way that alpha particles could be deflected at enormous angles as if they were impacting on a dense opaque region, and this dense region was located at the center of each atom.
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lithium reacts spontaneously with bromine to produce lithium bromide. if 25.0 g of lithum and 25.0 g of bromine are present at the beginning of the reaction how much of each substance are present when the reaction is complete
Answers
The balanced equation for the reaction of lithium and bromine to produce lithium bromide is:
Li + Br2 → LiBr
The total mass of the reactants and products must be equal in order for the rule of conservation of mass to apply. As a result, we can determine the mass of the other reactant and product if we know the mass of one reactant and one product.
In this instance, we are aware that the initial concentrations of the reactants are 25.0 g of lithium and 25.0 g of bromine. Lithium has a molar mass of 6.939 g/mol, while bromine has a molar mass of 159.808 g/mol. So that we may convert each substance's mass to a unit of moles:
Li: 25.0 g / 6.939 g/mol = 3.6 moles
Br2: 25.0 g / 159.808 g/mol = 0.156 moles
We can then convert the moles back to grams using the molar mass of each substance:
Li: 3.6 moles x 6.939 g/mol = 24.96 g
LiBr: 3.6 moles x (6.939 g/mol + 159.808 g/mol) = 627.38 g
Br2: 0.156 moles x 159.808 g/mol = 24.96 g
So, when the reaction is complete, we have 24.96 g of lithium, 24.96 g of bromine, and 627.38 g of lithium bromide.
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What is the molar mass of Cu2CO3? You may round your answer to the nearest hundredth. Be sure to place the correct units on the end
Answers
Molar mass of Cu2CO3 is 187.05 g/mol
Molar mass M is the product of the mass of a chemical compound and the molecular weight of the substance. MB = m/nB, where m is the total mass of a sample of a pure material and nB is the molecular weight of substance B, is the formula for it. Pure substance is covered under the definition. Any material has a molecular weight of 6.023 x 1023. (Avagadro number). It can be used to quantify the results of a chemical reaction. Mol is used to indicate the unit.
= 63.5*2+12.01+16*3
= 187.05
so the molar mass of Cu2Co3 is 187.05
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Three magnesium isotopes have atomic masses and relative abundances of 23.995 amu (78.99%), 24.986 amu (10.00%), and 25.992 (11.01%). Calculate the average atomic mass of magnesium.
i’ll give brainliest
Answers
You would multiple the percentages times the atomic mass and then add each of them together.
Ex. (23.995)(.7899)+(24.986)(.10)+(25.992)(.1101). = 24.3139 amu
what 3 molecules are made from the process of cellular respiration?
Answers
Answer:
Hydrogen (H20), Carbon Dioxide (CO2), ATP
The percentage of oxygen in air is 21%.
The mass of air in a classroom was 220Kg
Calculate the mass of oxygen in the classroom
Answers
The mass of oxygen in the room would be 46.2 kg.
The mass of a component as a percentage of the mass of the substance itself is known as the percent composition. Mathematically, this is expressed as:
Mass of component/mass of material multiplied by 100% gives the percent composition.
The percentage of oxygen in the air in this instance is represented as 21%, while the total mass of air in the classroom is provided as 220 kg. This formula can be used to determine how much oxygen is present in the classroom:
% oxygen composition = mass of oxygen/mass of air
Making the oxygen mass the focus of the formula:
Mass of air x percent oxygen content equals mass of oxygen
Let's replace the values now:
Weight of oxygen = 21% times 220
= 0.21 x 220
= 46.2 kg
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What number of ATP is produced when a molecule of glucose undergoes fermentation?
A. 4
B. 36
C. 2
D. 38
Answers
C. 2 ATP. At the cellular level, energy is used and stored as adenosine triphosphate (ATP). Adenine, ribose, and three serially bound phosphate groups make up.
the structure of ATP, which is a nucleoside triphosphate. An organic substance called adenosine triphosphate provides energy for a variety of biological functions in living cells, including muscular contraction, nerve impulse transmission, condensate dissolving, and chemical synthesis. The main molecule for storing and transmitting energy in cells is adenosine 5'-triphosphate (ATP). At the cellular level, energy is used and stored as adenosine triphosphate (ATP). Adenine, ribose, and three serially bound phosphate groups make up. It is frequently referred to as the cell's energy currency and is like the money that is kept in a bank.provides energy for a variety of biological functions in living cells, including muscular contraction, nerve impulse transmission,
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what is the age of a sample of charcoal in which 90% of the carbon-14 that was originally present has decayed? assume the half-life of carbon-14 isotope is 5700 years
Answers
The age of a sample of charcoal in which 90% of the carbon-14 that was originally present has decayed is about 18,935 years.
A = A₀ (1/2)ᵗ/ᵏ (1) Half-life formula.
0.1A₀ = A₀ (1/2)ᵗ/⁵⁷⁰⁰ (2)Substitute A=0.1A₀ and k=5700
0.1 = (1/2)ᵗ/⁵⁷⁰⁰ (3)Divide both sides by A₀.
log₁/₂0.1 = t/5700 (4)Convert to log from using x=bʸ - logbx = y
5700log₁/₂0.1 = t (5)Multiply both sides by 5700.
18,935 = t
Use the half-life formula A = A₀(1/2)ᵗ/ᵏ where A is the initial amount, t is the number of years, and k is the half-life of the substance. It is given that 90% has decayed which means 10% remains so let A = 0.1A₀. Carbon-14 has a half-life of 5700 years so let k=5700. Substitute these values into the formula and solved for t to get 18,935 years as its value.
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HELP PLEASE BOND ENTHALPY WORKSHEET
Answers
A sample of brass released 2,690 J into some water. The initial and final temperature of the water are 21.0 °C and 42.0 °C, respectively. What is the mass of the water?
Answers
The mass of the water can be obtained as 31 g.
What is the mass of the water?
I would want us to have the first law of thermodynamics at the back of our minds. I said this because the energy that have been lost by the brass is the energy that the water is going to have to gain and this is the first law of thermodynamics.
Thus we have;
H = mcdT
m = mass of the water
c = Specific heat capacity of water
dT = temperature change
Then we have;
2690 = m * 4.18 * (42 - 21)
m = 2690/4.18 * (42 - 21)
m = 31 g
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How many grams of oxygen are in 125 grams of FeCO3?
Answers
Answer:
In 125 grams of feCO3 287.30 g of FeCO3 oxygen is produce
Explanation:
The Balance Chemical Equation is as follow,
FeCl₂ + Na₂CO₃ → FeCO₃ + 2 NaCl
Calculate Mass of FeCl₂ as,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 2 mol.L⁻¹ × 1.24 L
Moles = 2.48 mol
Also,
Moles = Mass ÷ M.Mass
Solving for Mass,
Mass = Moles × M.Mass
Putting Values,
Mass = 2.48 mol × 126.75 g.mol⁻¹
Mass = 314.34 g of FeCl2
Calculate Mass of FeCO₃ formed as,
According to equation,
126.75 g (1 mole) FeClZ produces = 115.85 g (1 mole) FeCO₃
So,
314.34 g of FeCl₂ will produce = X g of FeCO₃
Solving for X,
X = (314.34 g × 115.85 g) ÷ 126.75 g
X = 287.30 g of FeCO₃
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What element's don't have oxidation numbers?
Alkaline earth metals
noble gases
alkali metals
halogens
Answers
Noble gases does not have any oxidation number. And elements in pure form does not have any oxidation numbers.
Elements in their pure form, such as gold (Au), do not have oxidation numbers because they are not chemically bonded to any other elements. Some other examples of elements that do not have oxidation numbers include carbon (C) in a diamond, and hydrogen (H) in molecular hydrogen (H2).
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What is the relationship between a chromosome and DNA?
Responses
Chromosomes manufacture DNA.
Chromosomes are made of DNA.
DNA is made of chromosomes.
DNA manufactures chromosomes.
Answers
Answer:
I think it is b chromosomes are made of DNA. I'm not for sure tho sorry
HELP PLS I NEED ANSWER NOW PLS :[ Which substances are released during cellular respiration? Use complete sentences to explain how the mass of oxygen is conserved during cellular respiration.
Answers
consider the energy diagram for a solution formed between molecules a and b. what can be said about this solution?
Answers
The energy diagram for a solution formed between molecules A and B indicates that the solution is energetically favorable.
This means that when A and B are mixed together, the total energy of the system is lower than if they were kept separate.
This is due to the molecules forming strong intermolecular forces, such as hydrogen bonds, that make the solution more stable. Additionally, this type of solution can be described as being “miscible”, meaning that the two molecules are able to mix together to form a homogenous solution.
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which statement supports the main ideas of the law of conservation of mass? responses the reactants in a chemical reaction are the same as the products. the reactants in a chemical reaction are the same as the products. the masses of the reactants are less than the masses of the products. the masses of the reactants are less than the masses of the products. the masses of the reactants are equal to the masses of the products. the masses of the reactants are equal to the masses of the products. the masses of the reactants are greater than the masses of the products.
Answers
Answer:
the first statement is correct...
it only supports the main idea of the law of conservation of mass...
hope I can help you...
A closed container is filled with oxygen. The pressure in the container is 355 kPa What is the pressure in
millimeters of mercury?
Answers
A puddle of liquid ammonia is evaporating. What mass of the ammonia evaporates as it absorbs 8. 2 kilo joules of energy? Pls help
Answers
24g of mass of the ammonia evaporates as it absorbs 8. 2 kilo joules of energy.
q = m .ΔHf
[tex]8.2KJ=m(399)\frac{J}{g}\\\frac {8.2*10^{3} J}{339\frac{J}{g}}=m\\0.024*10^{3}g=m\\24g=m[/tex]
With the formula NH3, ammonia is a nitrogen and hydrogen inorganic chemical. Ammonia, the simplest pnictogen hydride and a stable binary hydride, is a colourless gas with a strong, pungent odour.When molecules on a liquid's surface escape into the vapour phase, evaporation takes place. When clusters of liquid molecules erupt into vapour bubbles, boiling happens within the liquid's bulk rather than at its surface. Below, at, and above their boiling points, all liquids will evaporate.It contributes considerably to the nutritional demands of terrestrial creatures by serving as a precursor to 45% of the world's food and fertilisers. Biologically, it is a frequent nitrogenous waste, especially among aquatic animals.
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Cyclopropane rearranges to form propene in the gas phase. The reaction is first order in cyclopropane and has a measured rate constant of k=3. 36×10^−5 s^−1 k=3. 36×10^−5 s^−1 at 720 K. If the initial cyclopropane concentration is 0. 0445 MM, what will the cyclopropane concentration be after 235. 0 min? Express the molarity to three significant figures
Answers
2.8×10−2 M will the cyclopropane concentration be after 235. 0 min.
The cycloalkane cyclopropane has the chemical formula (CH2)3, and it is made up of three methylene groups (CH2) that are joined together to create a ring. The structure experiences significant ring strain as a result of the ring's modest size.
[tex]k=3.36*{10^{-5}}{s^{-1}}\\\\K=\frac {2.303 }{t} log \frac{a}{a-x}\\\\3.36*{10^{-5}}=\frac {2.303 }{235min} log \frac{0.445}{a-x}\\\\\\3.36*{10^{-5}}=\frac {2.303 }{235*60} log \frac{0.445}{a-x}\\\\\\20.571*{10^{-2}}=log \frac{0.445}{a-x}\\\\\\1.58=\frac{0.445}{a-x}\\\\\\(a-x)=0.281\\\\(a-x)=2.8*{10^{-2}}M[/tex]
This meant that the quick and painless induction of anaesthesia using cyclopropane and oxygen was possible Ralph Waters, an American anaesthetist, developed cyclopropane and put it to use in medicine. To save this then-expensive substance, he devised a closed system with carbon dioxide absorption. With a blood/gas partition value of 0.55 and a minimum alveolar concentration of 17.5%, cyclopropane is a reasonably strong, non-irritating, and sweet-smelling substance. .8×10−2 M will the cyclopropane concentration be after 235. 0 min.
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