In the given question, Insulin, adrenaline, and estrogen are examples of hormones. Each hormone has a specific function and plays an important role in maintaining homeostasis in the body.
Insulin is a hormone made up of 51 amino acids and is released by the beta cells of pancreas. It helps in regulating blood glucose level.
Adrenaline is also known as epinephrine. It is produced by the adrenal glands and is involved in the body's "fight or flight" response to stress.
Estrogen is a female sex hormone produced by the ovaries and is involved in the development of female secondary sexual characteristics, such as breast development and the menstrual cycle.
Therefore, examples of hormones are insulin, adrenaline, and estrogen, which are chemical messengers that regulate many physiological processes in the body.
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Draw two lewis structures for a compound with the formula c4h10. no atom bears a charge, and all carbon atoms have complete octets.
The two Lewis structure for a compound with the formula [tex]\rm C_4H_{10}[/tex] are shown below in the attached image.
In a Lewis structure, an element's chemical symbol shows its nucleus and inner electrons, while dots or lines represent valence electrons. Valence electrons are the atom's outermost electrons and are important in chemical bonding.
Shown below are the two Lewis structures of the compound with molecular formula [tex]\rm C_4H_{10}[/tex]. The two structures are the isomers of each other.The number of sigma bonds = 5, and number of pi bonds = 3 in the given compound.Therefore, the two structure of [tex]\rm C_4H_{10}[/tex] are isomers of each other (refer to the attached image).
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Compare the ways in which atoms combine to form molecules and compounds
The valence shell holds up to two electrons in which 2 elements? a. H and C b. H and Li c. H and He d. He and Be
Answer : The correct option is, (d) He and Be
Explanation :
First we have to determine the electronic configuration of the following elements.
The electronic configuration of hydrogen (H) is:
[tex]1s^1[/tex]
The electronic configuration of carbon (C) is:
[tex]1s^22s^22p^2[/tex]
The electronic configuration of lithium (Li) is:
[tex]1s^22s^1[/tex]
The electronic configuration of helium (He) is:
[tex]1s^2[/tex]
The electronic configuration of beryllium (Be) is:
[tex]1s^22s^2[/tex]
From the electronic configuration of the elements we conclude that, the hydrogen element has '1' valence electron, carbon element has '4' valence electrons, lithium element has '1' valence electron, helium element has '2' valence electrons and beryllium element has '2' valence electrons.
Thus, the helium and beryllium are the elements that holds up to two electrons in their outermost shell.
Hence, the correct option is, (d) He and Be
Table salt contains 39.33 g of sodium per 100 g of salt. The U.S. Food and Drug Administration (FDA) recommends that adults consume less than 2.40 g of sodium per day. A particular snack mix contains 1.24 g of salt per 100 g of the mix.What mass of the snack mix can you consume and still be within the FDA limit?
The answer is: mass of the snack mix can you consume and still be within the FDA limit is 491 grams.
ω(Na) = m(Na) ÷ m(salt).
ω(Na) = 39.33 g ÷ 100 g.
ω(Na) = 0.3933; mass percentage of sodium in salt.
m(salt) = 2.4 g ÷ 0.3933.
m(salt) = 6.10 g; mass of salt recommended for consumation.
Make proportion: 1.24 g : 100 g = 6.10 g ÷ m(mix).
m(mix) = 491.93 g; mass of the snack mix.
A transmission electron microscope was used to examine a microscopic organism. No nucleus was foumnd. Which of the following correctly describes the organism
Draw the structure of the 1 isomer of c7h16 that contains 3 methyl branches on the parent chain
Answer:
2,2,3-trimethyl-butane
Explanation:
Hello,
On the attached picture, you will find one isomer with chemical formula C₇H₁₆ which is 2,2,3-trimethyl-butane as it has the required three methyl brances on the parent chain, therefore, they must be arranged in the inner carbons otherwise, they will be considered as part of the parent chain.
Best regards.
Final answer:
The structure of the 1 isomer of C7H16 with three methyl branches on the parent chain can be represented as 2,2,3-trimethylbutane, which has a butane backbone with two methyl groups on the second carbon and one on the third.
Explanation:
The student asked to draw the structure of the 1 isomer of C7H16 that contains 3 methyl branches on the parent chain. For heptane (C7H16), one possible isomer with three methyl branches on the parent chain could be 2,2,3-trimethylbutane. In this structure, we have a four-carbon (butane) chain as the parent, with methyl groups attached to the second and third carbons (two on the second and one on the third).
Here's how you could write it out in steps:
Start with a straight chain of four carbons, which is your butane base.Attach two methyl groups to the second carbon on the chain.Attach one methyl group to the third carbon on the chain.An atom has three full orbitals in its second energy level. How many electrons are present in the second energy level of the atom
Answer:
6 electrons are in the third level if the atom.
When aqueous solutions of lead(ii) nitrate (pb(no3)2) and potassium phosphate (k3po4) are mixed, the products are solid lead(ii) phosphate and aqueous potassium nitrate. write the balanced chemical equation for this reaction. (include states-of-matter under the given conditions in your answer. use the lowest possible whole number coefficients.)?
Answer: The balanced chemical equation is written below.
Explanation:
Balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side will be equal to the total number of individual atoms on the product side. These equations follow law of conservation of mass.
The chemical equation for the reaction of lead (II) nitrate and potassium phosphate follows:
[tex]3Pb(NO_3)_2(aq.)+2K_3PO_4(aq.)\rightarrow Pb_3(PO_4)_2(s)+6KNO_3(aq.)[/tex]
By Stoichiometry of the reaction:
3 moles of aqueous lead (II) nitrate reacts with 2 moles of aqueous solution of potassium phosphate to produce 1 mole of solid lead (II) phosphate and 6 moles of aqueous solution of potassium nitrate.
Hence, the balanced chemical equation is written above.
According to the bohr model of the atom, which electron transition would correspond to the shortest wavelength line in the visible emission spectra for hydrogen? hints
n = 6 to n = 2
Further explanationFrom several sources, we have prepared the following answer choices:
A. n = 2 to n = 5
B. n = 6 to n = 4
C. n = 3 to n = 2
D. n = 6 to n = 2
We will determine which electron transition would correspond to the shortest wavelength line in the visible emission spectra for hydrogen.
The amount of energy released or absorbed by electrons when moving from n₁ level to n₂ level is equal to
[tex]\boxed{ \ \Delta E = -13.6 \Big( \frac{1}{n_2^2} - \frac{1}{n_1^2} \Big) \ }[/tex] in eV.
This energy difference is equal to [tex]\boxed{ \ hf = \frac{hc}{\lambda} \ }[/tex], where f and λ are the frequency and wavelength of the radiation emitted or absorbed.
Thus, the wavelength is inversely proportional to the energy difference from the electron transition. To get the shortest wavelength, it is determined by the largest ΔE.
From the formula above, we practically only need to calculate part [tex]\boxed{ \ \Big( \frac{1}{n_2^2} - \frac{1}{n_1^2} \Big) \ }[/tex] which is directly proportional to ΔE. Then from the results of the calculation of this section, we will get the shortest wavelength from the largest result..
A. n₁ = 2 to n₂ = 5
[tex]\boxed{ \ \Big( \frac{1}{5^2} - \frac{1}{2^2} \Big) \ }[/tex]
[tex]\boxed{ \ -\frac{21}{100} \ }[/tex]
By taking the absolute value, we get [tex]\boxed{ \ 0.210 \ }[/tex]
B. n₁ = 6 to n₂ = 4
[tex]\boxed{ \ \Big( \frac{1}{4^2} - \frac{1}{6^2} \Big) \ }[/tex]
[tex]\boxed{ \ \frac{5}{144} \ }[/tex]
We get [tex]\boxed{ \ 0.0347 \ }[/tex]
C. n₁ = 3 to n₂ = 2
[tex]\boxed{ \ \Big( \frac{1}{2^2} - \frac{1}{3^2} \Big) \ }[/tex]
[tex]\boxed{ \ \frac{5}{36} \ }[/tex]
We get [tex]\boxed{ \ 0.1389 \ }[/tex]
D. n₁ = 6 to n₂ = 2
[tex]\boxed{ \ \Big( \frac{1}{2^2} - \frac{1}{6^2} \Big) \ }[/tex]
[tex]\boxed{ \ \frac{2}{9} \ }[/tex]
We get [tex]\boxed{ \ 0.222 \ }[/tex]
The last calculation above shows the greatest results so that the shortest wavelength is undoubtedly gained from the electron transition n = 6 to n = 2.
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The electronic transition from [tex]\boxed{{\text{D}}{\text{. n}} = {\text{6 to n}} = {\text{2}}}[/tex] corresponds to the shortest wavelength.
Further explanation:
Rydberg equation describes the relation of wavelength of spectral line with the transition values. The expression for Rydberg equation is as follows:
[tex]\dfrac{1}{\lambda } = \left( {{{\text{R}}_{\text{H}}}} \right)\left( {\dfrac{1}{{{{\left( {{{\text{n}}_{\text{1}}}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)[/tex] …… (1)
Here,
[tex]\lambda[/tex] is the wavelength of spectral line
[tex]{{\text{R}}_{\text{H}}}[/tex] is Rydberg constant that has the value
[tex]{{\text{n}}_{\text{1}}}[/tex] and [tex]{{\text{n}}_{\text{2}}}[/tex] are the two positive integers, where .
Rearrange equation (1) to calculate .
[tex]\lambda = \dfrac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\dfrac{1}{{{{\left( {{{\text{n}}_1}} \right)}^2}}} - \dfrac{1}{{{{\left( {{{\text{n}}_{\text{2}}}} \right)}^2}}}} \right)}}[/tex] …… (2)
A. n = 2 to n = 5
Substitute 2 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 5 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).
[tex]\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 5 \right)}^2}}}} \right)}} \\&= 4.34 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}[/tex]
B. n = 6 to n = 4
Substitute 4 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 6 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).
[tex]\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left( {\frac{1}{{{{\left( 4 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}} \right)}}\\&= 2.63 \times {10^{ - 6}}{\text{ m}}\\\end{aligned}[/tex]
C. n = 3 to n = 2
Substitute 2 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 3 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).
[tex]\begin{aligned}\lambda &= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}}\right)\left( {\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 3 \right)}^2}}}} \right)}} \\ &= 6.56 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}[/tex]
D. n = 6 to n = 2
Substitute 2 for [tex]{{\text{n}}_{\text{1}}}[/tex] and 6 for [tex]{{\text{n}}_{\text{2}}}[/tex] in equation (2).
[tex]\begin{aligned}\lambda&= \frac{1}{{\left( {1.097 \times {{10}^7}{\text{ }}{{\text{m}}^{ - 1}}} \right)\left({\frac{1}{{{{\left( 2 \right)}^2}}} - \frac{1}{{{{\left( 6 \right)}^2}}}}\right)}} \\&= 4.10 \times {10^{ - 7}}{\text{ m}}\\\end{aligned}[/tex]
The value of [tex]\lambda[/tex] for transition from n = 6 to n = 2 is the least and therefore this transition corresponds to the shortest wavelength.
Learn more:
Ranking of elements according to their first ionization energy: https://brainly.com/question/1550767 Chemical equation representing the first ionization energy for lithium: https://brainly.com/question/5880605
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Atomic structure
Keywords: Rydberg constant, wavelength, n1, n2, positive integers, transition, 2, 6, 3, 5, transition, Rh, spectral line, shortest wavelength.
88.5 mol of P4O10 contains how many moles of P?
The number of molecules in 1.0 mole of H2O is _____ as in one mole of O2.
In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form water and released 1.715 × 103 kJ of heat. Calculate the corresponding mass change in this process. (1 J = 1 kg · m2/s2)
If a drop of blood is 0.05 mL, how many drops of blood are in a blood collection tube that holds 2 mL ?
Final answer:
To find out how many 0.05 mL drops of blood are in a 2 mL blood collection tube, divide the total volume by the volume of one drop. The calculation shows that there are 40 drops in a 2 mL tube.
Explanation:
To determine how many drops of blood are in a blood collection tube that holds 2 mL, we need to understand the relationship between the volume of the drops and the total volume that the tube can hold. Given that each drop of blood is 0.05 mL, we can calculate the number of drops in 2 mL by dividing the total volume by the volume of a single drop.
Here is the step-by-step calculation:
Determine the volume of one drop: 0.05 mL.
Determine the total volume of the collection tube: 2 mL.
Divide the total volume by the volume of one drop: 2 mL / 0.05 mL per drop.
Calculate the number of drops: 40 drops.
How many total atoms are in 0.250g of P2O5?
Explanation:
According to the mole concept, there are [tex]6.022 \times 10^{22}[/tex] atoms present.
It is given that mass is 0.250 g. And, number of moles are equal to mass divided by molar mass.
Mathematically, No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
As molar mass of [tex]P_{2}O_{5}[/tex] is 283.88 g/mol. Therefore, putting given values into the above formula as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{0.250 g}{283.88 g/mol}[/tex]
= 0.0008 mol
Hence, number of atoms present in 0.0008 mol are as follows.
[tex]0.0008 mol \times 6.022 \times 10^{22} atoms/mol[/tex]
= 0.0048 atoms
Thus, we can conclude that there are 0.0048 atoms in 0.250 g of [tex]P_{2}O_{5}[/tex].
How many picograms are in 1 Megagram?
Use the mass spectrum of rubidium to determine the atomic mass of rubidium.
The atomic mass of rubidium can be determined using the mass spectrum of rubidium obtained from a mass spectrometer.
Explanation:The atomic mass of rubidium can be determined using the mass spectrum of rubidium obtained from a mass spectrometer. In a mass spectrometer, a sample of rubidium is vaporized and exposed to high-energy electrons, causing the rubidium atoms to become charged ions. These ions are then accelerated into a magnetic field, and the extent to which they are deflected depends on their mass-to-charge ratios. By measuring the relative deflections of the ions and analyzing the mass spectrum, chemists can determine the mass of rubidium.
A dilute solution is prepared by transferring 40.00 ml of a 0.3433 m stock solution to a 750.0 ml volumetric flask and diluting to mark. what is the molarity of this dilute solution?
Name the compound P4O10
Which of the following is a heterogeneous mixture
Given 8.50 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?
Answer:
Amount of ethyl butyrate formed = 11.3 g
Explanation:
The reaction between butanoic acid (C3H7COOH) and excess ethanol (C2H5OH) is:
[tex]C3H7COOH + C2H5OH \rightarrow C3H7COOC2H5 + H2O[/tex]
Since ethanol is the excess reagent, the formation of the product i.e. ethyl butyrate is influenced by the amount of butanoic acid present
Based on the reaction stoichiometry:
1 mole of butanoic acid produces 1 mole of ethyl butyrate
Mass of butanoic acid = 8.50 g
Molar mass of butanoic acid = 88 g/mol
[tex]Moles\ of \ butanoic\ acid = \frac{Mass}{Molar\ mass} =\frac{8.50g}{88g/mol}=0.097 moles[/tex]
Moles of ethyl butyrate = 0.097
Molar mass of ethyl butyrate= 116 g/mol
[tex]Mass\ of\ ethyl\ butyrate= 0.097*116 = 11.3 g[/tex]
what is the density of an unknown if 7.82g of it occupies a volume of 3.63ml
You are on an alien planet where the names for substances and the units of measures are very unfamiliar.
Nonetheless, you obtain 29 quibs of a substance called skvarnick.
You can trade this skvarnick for gold coins, but the vendors all measure skvarnick in units of sleps; not quibs.
10 quibs is equal to 4 sleps.
If you have 29 quibs of skvarnick, how many sleps do you have?
Round your answer to the nearest tenth (one decimal place).
Answer:
If we have 29 quibs of skvarnick they will be equal to 11.6 sleps.
Explanation:
Mass of a substance called skvarnick = 29 quibs
10 quibs is equal to 4 sleps. This means that 1 quibs is equal to 0.4 sleps.
10 quibs = 4 sleps
1 quibs = [tex]\frac{4}{10} sleps = 0.4 sleps[/tex]
Then 29 quibs will be:
[tex]29 quibs=29\times 0.4 sleps =11.6 sleps[/tex]
If we have 29 quibs of skvarnick they will be equal to 11.6 sleps.
What are the necessary steps to prevent oxidation during high temperature
In 1928, 47.5 g of a new element was isolated from 660 kg of the ore molybdenite. the percent by mass of this element in the ore was:
To take the percent by mass of this element, we use the formula:
% mass = (mass of element / mass of ore) * 100%
% mass = (47.5 g / (660 kg * 1000 g / kg)) * 100*
% mass = 7.20 x 10^-3 %
Answer: The mass percent of element in the ore is 0.0072 %
Explanation:
To calculate the mass percentage of element in ore, we use the equation:
[tex]\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{Mass of ore}}\times 100[/tex]
Mass of element = 47.5 g
Mass of ore = 660 kg = 660000 g (Conversion factor: 1 kg = 1000 g)
Putting values in above equation, we get:
[tex]\text{Mass percent of element}=\frac{47.5g}{660000g}\times 100=0.0072\%[/tex]
Hence, the mass percent of element in the ore is 0.0072 %
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Which has the greater density? (Refer to "Densities of Common Substances" table.) air at sea level
air at 20 km altitude
Sodium hydroxide, NaOH; sodium phosphate, Na3PO4; and sodium nitrate, NaNO3, are all common chemicals used in cleanser formulation. Rank the compounds in order from largest mass percent of sodium to smallest mass percent of sodium.
How many grams of hydrogen gas can be obtained from 0.373 grams of magnesium? Note that the equation is NOT balanced.
Mg(s) + HCl(aq) → MgCl2(aq) + H2(g)
The first step to answering this item, is to balance first the chemical reaction. This is shown below.
Mg(s) + 2HCl (aq) --> MgCl₂(aq) + H₂(g)
HCl is multiplied by 2 to equalize the number of the hydrogen and chlorine.
The molar masses of magnesium and hydrogen gas are 24.3 g and 2 g, respectively. Using dimensional analysis and conversions, mass of hydrogen is calculated through the equation,
Mass of H₂ = (0.373 g Mg)(1 mol Mg/24.3 g Mg)(1 mol H₂/ 1 mol Mg)(2 g H₂/1 mol H₂)
Mass of hydrogen = 0.031 g of hydrogen
ANSWER: 0.031 g H₂
A 5.23 g sample of a metal occupies 0.27 ml. identify the density of the metal.
The density of a metal given a mass of 5.23 g and a volume of 0.27 mL is calculated using the formula density = mass / volume, resulting in a density of 19.37 g/mL, with the final answer limited to four significant figures.
Explanation:To identify the density of a metal given its mass and volume, you can use the formula for density, which is density = mass / volume. Given a 5.23 g sample of metal that occupies 0.27 mL, you first ensure the units are correct for density in g/mL. You then simply divide the mass by the volume.
So, density = 5.23 g / 0.27 mL = 19.37 g/mL.
This calculation gives us the density of the metal. Remember, when calculating density, it's important to ensure that your units are consistent, and for density, g/mL is a common unit. Additionally, when expressing your final answer, it matches the number of significant figures from the given data, in this case, being limited to four significant figures due to the precision of the volume measurement.
Without doing any calculations, determine which mass is closest to the atomic mass of carbon. without doing any calculations, determine which mass is closest to the atomic mass of carbon. 12.00 amu 12.50 amu 13.00 amu
Final answer:
The atomic mass closest to carbon without calculations is 12.00 amu since carbon's average atomic mass is listed as 12.011 amu on the periodic table, and it mostly consists of carbon-12.
Explanation:
The atomic mass of an element like carbon is the weighted average of the masses of its naturally occurring isotopes. Carbon is predominantly composed of carbon-12, which has an atomic mass of exactly 12 amu (atomic mass units), as this isotope makes up about 98.90% of naturally occurring carbon. Because of this, the average atomic mass of carbon is close to 12 amu, but slightly higher due to the presence of the carbon-13 isotope. The periodic table lists carbon's atomic mass as 12.011 amu, making 12.00 amu the mass closest to carbon's atomic mass without doing any calculations.
You have a racemic mixture of (+)-2-butanol and (-)-2-butanol. the (+) isomer rotates polarized light by +13.5∘. what is the observed rotation of your mixture?
Answer:
The degree of rotation of racemic mixture will equal 0°
Explanation:
Hello,
(-) isomer is in other words the Enantiomer of (+) isomer and as long as the statement indicates that (+)-isomer optical rotation is + 13.5°, the optical rotation of (-)-isomer will be the same value and degree but with the inverse sign which is -13.5°.
In such a way, the degree of rotation of the taken into account racemic mixture will equal 0°
This is in this way due to the fact that the racemic mixture has equal amount of both enantiomers.
Best regards.