Please comment the chart!!!!!
thanks!!

Answers

Answer 1

Answer:

Exothermihic chart

Explanation:


Related Questions

what is valency of an atom?​

Answers

The number of replaceable electrons in an atom is called its valency.

Examples

Monovalent - HydrogenDivalent - Oxygen

Valency = 8 - Number of electron in last shell [When number of electrons in last shell > 4]

Valency = Number of electron in last shell [When number of electrons in last shell < 4]

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Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

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

Answers

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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The decrease in the water table due to overuse of water.

Answers

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.

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state function and non state function ​

Answers

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:

how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

Answers

Answer:

how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

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.

Answers

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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cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?

Answers

1.37% of cesium–131 will remain 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 / 9.7

Finally, 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%

NOTE; N/N₀ is the fraction remaining.

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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

Answers

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.  

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 ​

Answers

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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Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?

Answers

cyclohexane will distill off first as it will have lower boiling point compared to ortho xylene which has higher molecular mass

11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride

Answers

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

calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume​

Answers

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:

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if wrong just correct me

Which subshells are found in each of the following shells
electron subshell - M shell

Answers

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.

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)

Answers

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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which of the following is indicated by the ph value of a solution?
a- it's hydrogen ion concentration
b- its ammonium ion concentration
c- ability to undergo chemical reaction
d- its ratio of solute amount to solvent volume

Answers

Answer:

c- ability to undergo chemical reaction

1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.

Answers

The structures are shown in the image attached.

A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.

Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.

I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2)  and tetrahydrofuran (image 3).

All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.

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name a factor tht affects the value of electron affinity​

Answers

Answer:

Atomic sizeNuclear chargesymmetry of the electronic configuration
Various factors that affect electron affinity are atomic size, nuclear charge and the symmetry of the electronic configuration. Atomic size: With increase in the atomic size, the distance between the nucleus and the incoming electron also increases.

Select the choice that best completes the following sentence: When cooled slowly, transformations near the melting temperature tend to yield ______ grains due to the formation of ______ nucleation sites followed by ______ grain growth.

Answers

Question Completion with Options:

O coarse...few...rapid

O fine...few...slow

O fine...multiple...rapid

O coarse...few...slow

O fine...multiple...slow

Answer:

The choice that best completes the sentence is:

O coarse...few...slow

Explanation:

Transformations near the melting temperature develop coarse grains because few nucleation sites are formed and the rate of the grain growth is usually slow.  This is because of the process that starts with  recrystallization, recovery, and nucleation before growth can occur.  While recrystallization enables the grain to increase in size at high temperature, nucleation gives the grain the energy to irreversibly grow into larger-sized nucleus.

Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.

Answers

Answer:

10.77%

Explanation:

Molar mass of Cu = mass deposited/number of moles of Cu

Molar mass of Cu = 0.4391 g/6.238x10^-3 moles

Molar mass of Cu = 70.391 g/mol

%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100

%error = 10.77%

what is the difference between 25ml and 25.00ml​

Answers

Answer:

There is no difference between the two.

Explanation:

They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments

what are the properety of covalent bond​

Answers

Explanation:

1. boiling and melting point

2. electrical conductivity

3. Bond strength

4. bond length

A covalent bond consists of negative electrons that are shared in between atoms. Because of this bond, they possess and manifest physical abilities, including electrical pressure/conductivity and lower melting points compared to ionic compounds.

Considering a fish breeder decided to breed small fishes which needs a pH between 6,0 to 7,0 to stay alive. He needs to adjust the water's pH that is 5,0 to a value of 6.5, having available only calcium carbonate. The mass in mg added to 5L of water is about:



A)2,5

B)5,5

C)6,5

D)7,5

E)9,5

Answers

6.5< x < 8.5 hope this helps

What type of bonding is occuring in the compound below?

A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar

Answers

Answer:

(B). it's metallic bonding

If the starting material has no stereogenic centers, when carbonyl compounds are reduced with a reagent such as LiAlH4 or NaBH4 and a new stereogenic center is formed, what will the composition of the product mixture be?
A) Forms a racemic mixture of the two possible enantiomers.
B) Forms more of one enantiomer than another because of steric reasons around the carbonyl.
C) Forms more of one enantiomer than another depending on the temperature of the reaction.
D) Forms different products depending on the solvent used.

Answers

Answer:

A) Forms a racemic mixture of the two possible enantiomers

When carbonyl compounds are reduced with a reagent such as LiAlH₄ or NaBH₄ and  new stereogenic center is formed chemical change will lead to products that form a racemic mixture of the two possible enantiomers.

What is a chemical change?

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

There are three types of chemical changes:

1) inorganic changes

2)organic changes

3) biochemical changes

During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.

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tea contains approximately 2% caffeine by weight. assuming that you started with 18g of tea leaves, calculate your percent yield of extraced caffeine

Answers

.36 g of caffeine for this problem. 2% of 18g is 0.36g

How do I do this? What are the answers to the 5 questions shown?

Answers

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb

Answers

Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

1.8×10⁻⁶ = x²

√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺]7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH

Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 39.7 mg of this isotope, what mass remains after 48.2 days have passed?

Answers

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

[A] = 11.9g remains after 48.2 days

A 12.37 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.197 g. Identify the empirical formula of the new oxide

Answers

Answer:

MoO2

Explanation:

The empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of Mo2O3. Twice these moles = Moles of Mo. With the moles of Mo we can find its mass.

The difference in masses between mass of new oxide and mass of Mo = Mass of oxygen. With the mass of oxygen we can find its moles and the empirical formula as follows:

Moles Mo2O3 -Molar mass: 239.9g/mol-

12.37g * (1mol / 239.9g) = 0.05156 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.1031 moles of Mo

Mass Mo -95.95g/mol-:

0.1031 moles of Mo * (95.95g/mol) = 9.895g of Mo

Mass oxygen in the oxide:

13.197 - 9.895g = 3.302g Oxygen

Moles oxygen -Molar mass: 16g/mol-:

3.302g Oxygen * (1mol / 16g) = 0.206 moles O

Now, the ratio of moles O / moles Mo is:

0.206 moles O / 0.1031 moles Mo = 2

That means there are 2 moles of O per mole of Mo and the empirical formula of the new oxide is:

MoO2

here is the question

Answers

Answer:

1. Nitrate ions, NaNO3 - Sodium nitrate.

2. Sulphide ions, K2S - Potassium sulphide.

3. Sulphate ions, CaSO4 - Calcium sulphate.

4. Hydrogensulphite ions, NaHSO3 - Sodium hydrogensulphite.

5. Carbonate ions, CaCO3 - Calcium carbonate.

6. Hydrogencarbonate ions, KHCO3 - Potassium hydrogencarbonate.

7. Phosphite ions, PH3 - Hydrogen phosphite.

8. Nitride ions, NH3 - Hydrogen nitride ( ammonia ).

9. Ethanoate ions, CH3COONa - Sodium ethanoate.

10. Methanoate ions, HCOONa - Sodium methanoate.

11. Fluoride ions, HF - Hydrogen fluoride.

12. Chloride ions, KCl - Potassium chloride.

13. Bromide ions, HBr - Hydrogen bromide.

14. Iodide ions, NaI - Sodium iodide.

15. Phosphate ions, K3PO3 - potassium phosphate.

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