Complete question is;
Identify the type of reaction in the chemical reaction below:
2P205 ➡️ 4P + 502
single replacement
synthesis
decomposition
combustion
double replacement
Answer:
Decomposition
Explanation:
We. An see in the question that the compound 2P205 is broken down into simpler substances which are phosphorus (P) and oxygen (O).
Now, this is a decomposition reaction because a decomposition reaction is one in which a compound is broken down into simpler substances
Starting from (R)-3-methylhex-1-yne as the substrate at the center of your page, draw a reaction map showing the regiochemical and stereochemical outcome or outcomes for each of the following series of reagents. Name each of your products, including stereochemical designations for any chirality centers that are generated.
a. HgSO4, H2SO4, H2O
b. 1. 9-BBN; 2. H2O2, NaOH
c. Br2, CCl4
d. HBr
Solution :
A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.
Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.
In the context, the products including the stereochemical designations for any chirality centers starting from the (R)-3-methylhex-1-yne as the substrate are attached below.
Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution.
Answer:
Molarity is 0.075 M.
Explanation:
Moles:
[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]
RFM of potassium sulphate :
[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]
substitute:
[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]
In volume of 5.00 l:
[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]
Which subshells are found in each of the following shells
electron subshell - M shell
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?
11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)
The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.
How to calculate moles in stoichiometry?Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.
According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:
Mg + Cl₂ → MgCl₂
Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.
This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.
Next, we convert moles of magnesium chloride to mass as follows:
molar mass of magnesium chloride = 95.211g/mol
mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.
Therefore, 218.99 grams of magnesium chloride will be formed.
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#SPJ1
cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Percentage remaining after 60 days =?Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 9.7 days
Time (t) = 60 days
Number of half-lives (n) =?
n = t / t½
n = 60 / 9.7Finally, we shall determine the percentage remaining. This can be obtained as follow:
Let the original amount be N₀
Let the amount remaining be N
Number of half-lives (n) = 60 / 9.7
N = N₀ / 2ⁿ
Divide both side by N₀
N/N₀ = 1/2ⁿ
N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾
N/N₀ = 0.0137
Multiply by 100 to express in percentage
N/N₀ = 0.0137 × 100
N/N₀ = 1.37%Therefore, the percentage remaining after 60 days is 1.37%
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Calculate the pH of each solution.
A. 0.18 M CH3NH2
B. 0.18 M CH3NH3Cl
C. a mixture of 0.18 M CH3NH2 and 0.18 M CH3NH3Cl
Answer:
See Explanations
Explanation:
pH =-log[H₃O⁺] = -log[H⁺]
pOH = -log[OH⁻]
For weak acids [H⁺] = SqrRt(Ka·[Acid])
For weak bases [OH⁻] = SqrRt(Kb·[Base])
pH + pOH = 14
__________________________________________
A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95
CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;
[OH⁻] = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M
=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05
=> pH = 14 - pOH = 14 - 2.05 = 11.95.
*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.
___________________________________________________
B. Given 0.18M CH₃NH₃Cl
In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴
Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.
Hydrolysis Reaction of Methylammonium Ion:
CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻
Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹ Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.
*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.
C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl
Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl
In Water ...
=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl
=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻
=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻
-----------------------------------------------------------
Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)
=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹
----------------------------------------------------------
From the 0.36M CH₃NH₃⁺
=> CH₃NH₃⁺ + H₂O ⇄ CH₃NH₄⁺ + OH⁻
C(eq) 0.36M ---- x x (<= at equilibrium after mixing)
Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)
=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M
=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process
=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻
--------------------------------------------------------
Cl⁻ + H₂O => No Rxn (Cl⁻ does not react with H₂O)The 0.1926M in OH⁻ => [H⁺] = Kw/[OH⁻] = (10⁻¹⁴/0.1926)M = 5.192 x 10⁻¹⁴M in H₃O⁺ ions (= H⁺ ions) ...∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix
The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and
CH₃NH₃Cl and the mixture give the following approximate values;
A. The pH value of the 0.18 M CH₃NH₂ is 11.93
B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69
C. The pH value of the mixture is 10.644
Which method can be used to calculate the pH values?A. 0.18 M CH₃NH₂
The solution is presented as follows;
CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻
Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we
have;
The number of moles of CH₃NH₂ remaining = 0.18 - x
Which gives;
[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]
[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴
Therefore;
[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]
4.167 × 10⁻⁴ × (0.18 - x) = x²
4.167 × 10⁻⁴ × (0.18 - x) - x² = 0
Which gives;
x = [OH⁻] = 8.455 × 10⁻³
pH = 14 + log[OH⁻]
Which gives;
pH = 14 + log(8.455 × 10⁻³) ≈ 11.93
B. 0.18 M CH₃NH₃Cl
The solution is presented as follows;
CH₃NH₃⁺ → CH₃NH₂ + H⁺
Let x represent the number of moles of CH₃NH₂ and H⁺ produced,
respectively, we have;
The number of moles of CH₃NH₃⁺ remaining = 0.18 - x
Which gives;
[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]
Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹
Therefore;
[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]
2.27 × 10⁻¹¹ × (0.18 - x) = x²
2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0
Which gives;
x = [H⁺] ≈ 2.02 × 10⁻⁶
pH = -log[H⁺]
Which gives;
pH = -log(2.02 × 10⁻⁶) ≈ 5.69
C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;
Based on the Henderson-Hasselbalch equation, we have;
[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]
Which gives;
[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]
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For the following acids of varying concentrations, which are titrated with 0.50 M NaOH, rank the acids in order of least to most volume of base needed to completely neutralize the acid.
a. 0.2M H2C6H5O7
b. 0.2M H2C2O4
Answer:
0.2M H2C6H5O7 < 0.2M H2C2O4
Explanation:
A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.
More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.
Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.
H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.
H₂C₆H₅O₇ < H₂C₂O₄
The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)
Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.
Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.
Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.
Strong acids ionizes completely to produce hydrogen ions.
Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.
In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.
H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23
H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08
Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇
For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.
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what type of bonding does Sodium Sulphate comes under?and explain in detail please
Answer:
The bond between sodium sulfate is an ionic bond since it's a bond between a metal and non metals however the bond between sulfur and oxygen is a covalent bond since the two are non metals and the other reason that makes this an ionic bond is that there is both losing and gaining of electrons..
I hope this helps
Van der Waals forces hold molecules together by: A. moving electrons from one molecule to another. B. attracting a lone pair of electrons to the positive charge of a hydrogen. C. inducing temporary dipoles that attract each other. D. sharing electrons between atoms.
Van der Waals forces hold molecules together by inducing temporary dipoles that attract each other. That is option C
Van Der Waals forces are example of those intermolecular forces which are weaker than ionic and covalent bonds that exists between molecules.
Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the of real gases at low temperatures and high pressures that:
electrons in a non-polar molecule such as hydrogen are close to one nucleus as to the other, although momentary concentration at one end of the molecule may occur, this momentary concentration of electron cloud on one side create a temporary dipole in the hydrogen molecule, that is, one side of the molecule acquires a partial negative charge while the other side acquires a partial positive charge of equal magnitude, the temporary dipole induces a similar dipole in an adjacent behavior molecule, this results in a temporary dipole-induced dipole attraction between the positive and negative ends of the adjacent molecules.This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT
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Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.
Answer:
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
Explanation:
The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.
The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.
The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.
From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.
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Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
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Which phenomenon explained below is an example of deposition?
Select the correct answer below:
A) Hail is formed from water droplets lifted by air currents to an altitude where they turn into pellets of ice.
B) Frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
C) In the winter, the top few inches of a pond turn to ice.
D) The visible cloud arising from a boiling tea kettle is not actually steam, but droplets of liquid water that form as the
steam cools in the air.
Answer:
b
Explanation:
deposition is when water turns from gas to solid. b is the only one that fits
Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
What is deposition?Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.
Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.
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The decrease in the water table due to overuse of water.
Answer: Groundwater and surface water are connected. When groundwater is overused, the lakes, streams, and rivers connected to groundwater can also have their supply diminished. Land subsidence occurs when there is a loss of support below ground. This is most often caused by human activities, mainly from the overuse of groundwater, when the soil collapses, compacts, and drops.
Hope it Helps. :)
Compare the modern (electron cloud) model of the atom with Bohr’s atomic model. Which of these statements describe the two models correctly? Check all of the boxes that apply.
A. Bohr’s model was replaced only because of its age.
B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model the location of the electrons cannot be predicted.
C. The modern model explains all available data about atoms; Bohr’s model does not.
D. The modern model is more widely accepted because it was proposed by more well known scientists.
Answer:
B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model, the location of the electrons cannot be predicted.
AND
C. The modern model explains all available data about atoms; Bohr’s model does not.
Explanation:
The answers are right on Edge. :)
Answer:
b and c
Explanation:
my assignment was 100% 2022
Kevin's supervisor, Jill, has asked for an update on today's sales, Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her ? a) Call with a quick update Ob ) Send a detailed text message c ) Book a one-hour meeting for tomorrow morning d) Send a detailed email
Answer:
d
Explanation:
since it is much convenient since the email will not get lost and it's contents will not be forgotten
Which one of the following is not matches the organelle with its function
Answer:
rip there isnt a photo
Explanation:
i do know a lot about cells tho lol
An antacid tablet weighing 1.30g was fully neutralized at 42.00 mL(an excess amount) of 0.250MHCl. 10.00 mL of 0.100 M NaOH was then used to back titrate the excess HCl. How many moles of acid did the antacid neutralize
Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
Moles HCl added:
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
Moles NaOH to titrate the excess:
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
Moles of acid that were neutralized:
0.0105 moles - 0.0010 moles =
0.0095 moles of acid were neutralized by the antiacidSally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to one compartment of the microwell plate. She then makes a solution of MCln by adding KCl to M(NO3)n. She adds 7.903 mL of the resulting mixture to a second compartment of the microwell plate. Sally knows n = +2. She has already calculated [Mn+] in the prepared MCln solution using the Nernst equation. [Mn+] = 8.279 M
Required:
How many moles of [Cl-] must be dissolved in that compartment?
Answer:
0.1309 mol
Explanation:
From the given information:
The metal ion, two ions of [tex]M^{+}[/tex] reacted with Cl⁻ to form [tex]MCl_n[/tex] i.e. the compound formed is [tex]MCl_2[/tex].
The concentration of the metal ion formed [tex][M^+][/tex] = 8.279 M
The concentration of the chlorine ion formed [tex][Cl^-][/tex] = 2 × 8.279 M
= 16.558 M
∴
We know that:
[tex]\mathsf{Molarity = \dfrac{no \ of \ moles }{volume (mL)}}[/tex]
The number of moles of [tex][Cl^-][/tex] = [tex]16.558 \ mol.L^{-1} \times 7.903 \ mL \times \dfrac{1 \ L}{1000 \ mL}[/tex]
= 0.1309 mol
calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume
Answer :
volume of a gas = weight * 22.4 l / gram molecular weight
volume of o2 = ?
weight given = 20.5 g
gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )
volume of oxygen = 20.5 * 22.4 / 32
volume of oxygen = 14.35 liters
Explanation:
hope this helps you
if wrong just correct me
Determine the value of the equilibrium constant for the following reaction, where the following amounts of each species are present at equilibrium in a 5.00 L container: 1.34 mol HCl, 4.30 mol O2, 30 g H2O, and 2.42 mol Cl2.
4 HCl(g) O2(g) ----> 2 H2O(l) 2 Cl2(g)
Explanation:
here's the answer to your question about
1. A
a stiff structure that surrounds and
protects a coll: found in plant, fungus, and some bacteria cells.
2.
is living things consisting of many cells.
3.
a green pigment that traps energy
from the sun.
the process in which plants and
some other organisms use the energy in sunlight to make food.
5. A
found in the nucleus of a cell,
a long nucleic acid molecule containing the genetic instructions
for the development and functioning of all living organisms.
Answer:
1 cell wall
2 yes
3 chloroplast
4 photosynthesis
5 Deoxyribonucleic acid (I believe)
hope this helped a little and pls mark brainiest if it did :)
Explanation:
The cell wall is a rigid layer that is found outside the cell membrane and surrounds the cell, providing structural support and protection.
When comparing Be2 and H2:
I. Be2 is more stable because it contains both bonding and antibonding valence electrons.
II. H2 has a higher bond order than Be2.
III. H2 is more stable because it only contains 1s electrons.
IV. H2 is more stable because it is diamagnetic, whereas Be2 is paramagnetic
a. II,III,IV
b.II,III
c.III only
d.I,II
e.III,IV.
Answer:
The answer is "Option b".
Explanation:
H2 does have bond energy of 1, while Be2 has a covalent bond of zero. Be2 has eight electrons, each of which dwells in a distinct orbital. As just a result, four of them are linked molecular orbitals and two are antibonding molecular orbitals, respectively. As just a result, this molecule is unstable. This chemical orbital, with a bond order of 1, has just two electrons. As a result, it is a very solid substance. H2's bond length is higher than Be2's. Since it only has one electron, H2 is more stable than that of other compounds.
state function and non state function
Answer:
State functions represent quantities or properties of a thermodynamic system, while non-state functions represent a process during which the state functions change. For example, the state function PV is proportional to the internal energy of an ideal gas, but the work W is the amount of energy transferred as the system performs work.
Explanation:
Which redox reaction would most likely occur if silver and copper metal were added to a solution that contained silver and copper ions?
A. Cu + Agt Cu2+ + 2Ag
B. Cu2+ + 2Ag* → Cu + 2Ag
C. Cu2+ + 2Ag → Cu + 2Ag+
D. Cu + 2Ag Cu²+ + 2Ag+
give the wrong answer and I'm reporting
Answer:
B
Explanation:
b/c copper is readuction agent
The most likely redox reaction that would occur if silver and copper metal were added to a solution that contained silver and copper ions is [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]. The correct answer is option C.
Redox reaction is a reaction in which reduction and oxidation takes place simultaneously.
In this reaction:
[tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]
Copper metal has a higher reduction potential than silver metal, which means that it will be oxidized to [tex]\rm Cu^{2+}[/tex] ions before silver metal is oxidized to [tex]\rm Ag^+[/tex] ions.
The [tex]\rm Cu^{2+}[/tex] ions in the solution will then react with the silver metal to form [tex]\rm Ag^+[/tex] ions and Copper metal. This reaction is an example of a displacement reaction, where a more reactive metal removes a less reactive metal from its compound.
Therefore, option C. [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex] is the correct answer.
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the nutrition label on rice lists the amounts of protein, carbohydrates and fats in one serving. these substances are important for human nutrition
Answer:
Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.
Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.
Explanation:
how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?
Answer:
how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?
Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.
a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.
1. Increase
2. decrease
3. No effect
Answer:
a. Decrease
b. Increase
c. Increase
d. No effect
Explanation:
Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.
a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease
b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect
c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase
d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase
A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.
B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.
C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.
D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.
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2.5
Nakula investigated the effect of heat on amylase. Amylase is an enzyme
that makes starch molecules break down into sugar molecules.
P
• Nakula put some amylase solution into two boiling tubes,
P and Q
• He boiled the solution in tube P. He did not heat tube Q.
• He waited until the solution in tube P had cooled down to
room temperature.
• He added equal volumes of starch solution to tube P and
boiled amylase amylase
• After 10 minutes, he tested both tubes for sugar.
and starch and starch
• Nakula found that there was sugar in tube Q. but not in tube P.
The structure of amylase deteriorates due to high temperature of the solution.
This experiment shows that the structure of amylase deteriorates due to high temperature which prevents this amylase from performing its function properly.
At high temperatures the amylase will break starch down very slowly or not at all due to denaturation of the enzyme's active site due to which it can't perform its function properly so we can conclude that high temperature denatures amylase enzyme.
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Nakula's conclusion was- "My results show that boiling destroys amylase"
Amylase is an enzyme that breaks down starch molecules into sugar molecules. He boiled the solution in tube P, and when he checked tube P for sugar, there wasn't any. He didn't boil the solution in tube Q and he found sugar in it.
Amylase had broken down starch molecules to sugar molecules in tube Q. Tube P's solution had been boiled, and this showed that when he boiled it, it destroyed the amylase, that is why the starch molecules hadn't been broken down into sugar molecules.
