How many electrons are shown in the following electron
configuration: 1s22s22p63s 23p64s23d104p65s24d105p66s2 ?
Express your answer numerically as an integer.

Answer:

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

A cell is the structural and fundamental unit of life. The study of cells from its basic structure to the functions of every cell organelle is called Cell Biology. Robert Hooke was the first Biologist who discovered cells

two types of cell

1) Prokaryotes

2) Eukaryotes

Characteristics of Cells

1) Cells provide structure and support to the body of an organism.

2) The cell interior is organised into different individual organelles surrounded by a separate membrane.

3) The nucleus (major organelle) holds genetic information necessary for reproduction and cell growth

[tex]hope \: its \: helpful \: to \: you \: please \: mark \: me \: a \: brainliest[/tex]

A cell is defined as the fundamental, structural and functional unit of all life.

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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

Answer:

Both have solutions in the  graduated cylinder.

Explanation:

Recording the amount of alcohol used as the difference between two masses is the wrong method used for measuring tert-butyl alcohol for the experiment.  For measuring tert-butyl alcohol for this experiment, the student has to measure the two masses when both the graduated cylinders has solution of tert-butyl alcohol not when one of it is empty (having no tert-butyl alcohol ).

The wrong aspect is that the liquid didn't need to be weighed. Instead the volume should have been recorded with the aid of the graduated cylinder.

This is a cylinder with marked readings and is used to measure the volume of liquids in the laboratory.

The graduated cylinder will accurately measure the amount of  alcohol used due to it being volatile and the mass fluctuating during the measurement.

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The concentration of CO₃²⁻ ions will be higher

To explain, I want you to imagine H₂CO₃ in water.

we know that it will lose 2 of it's protons, and form 2 ions

The ion which is more stable will have a higher concentration because that ion will refuse to react with anything else, so once something turns into that specific ion, it's not going back... unless there's a more stabler ion possible

In this case, the 2 ions formed are: HCO₃⁻ and CO₃⁽²⁻⁾, drawing the structures of both the ions tells us that both of them have resonance, but the CO₃⁽²⁻⁾ ion has more resonance structures and hence is more stable

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

Answer:

[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]

Explanation:

We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).

To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the individual elements in the compound: calcium and chloride.

Notice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.

We will convert using dimensional analysis, so we must create a ratio using the molar mass.

[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.

[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

Flip the ratio so the units of grams of calcium chloride cancel.

[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]

[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]

[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]

[tex]0.1351594882\ mol \ CaCl_2[/tex]

The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.

[tex]0.14 \ mol \ CaCl_2[/tex]

15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.

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

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.

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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Answer:

The degree of compactness of a substance

Answer:

D. -80m/s^2

Explanation:

V = u + at

5 = 65 + a (0.75)

0.75a = -60

a = -60/0.75

a = -80m/s^2

Therefore, is decelerating at 80m/s^2

Answer:

[tex]\boxed {\boxed {\sf D. \ 80 \ m/s^2 \ down}}[/tex]

Explanation:

We are asked to find the acceleration of a skydiver. Acceleration is the change in velocity over the change in time, so the formula for calculating acceleration is:

[tex]a= \frac{v_f-v_i}{t}[/tex]

The skydiver was initially traveling 65 meters per second, then he slowed down to a final velocity of 5 meters per second. He slowed down in 0.75 seconds.

[tex]\bullet \ v_f = 5 \ m/s \\\bullet \ v_i= 65 \ m/s \\\bullet \ t= 0.75 \ s[/tex]

Substitute the values into the formula.

[tex]a= \frac{ 5 \ m/s - 65 \ m/s}{0.75 \ s}[/tex]

Solve the numerator.

[tex]a= \frac{-60 \ m/s}{0.75 \ s}[/tex]

Divide.

[tex]a= -80 \ m/s^2[/tex]

The acceleration of the skydiver is -80 meters per second squared or 80 meters per second squared down. The skydiver is slowing down or decelerating, so the acceleration is negative or down.

Explanation:

here's the answer to your question

In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.

The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:

The addition of barium hydroxide will raise the pH slightly because the buffer still working.

The initial moles of those species are:

Hypochlorous acid:

[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]

Sodium hypochlorite:

[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]

Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:

Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O

For a complete reaction of 0.092 moles of barium hydroxide are required:

[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]

As there are 0.370 moles, the moles of HClO after the reaction are:

0.370 moles - 0.184 moles = 0.186 moles of HClO will remain

As you still have hypochlorite and hypochlorous acid you still have a buffer.

Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.

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Answer:

c. C3H9N2O2

Explanation:

The empirical formula of a compound is defined as the simplest whole number ratio of atoms present in a molecule. To solve this question we need to convert the mass of each atom to moles. With the moles we can find the ratio as follows:

Moles N -Molar mass: 14.01g/mol-

0.420g N * (1mol/14.01g) = 0.0300 moles N

Moles O -Molar mass: 16g/mol-

0.480g O * (1mol/16g) = 0.0300 moles O

Moles C -Molar mass: 12.01g/mol-

0.540g C * (1mol/12.01g) = 0.0450 moles C

Moles H -Molar mass: 1.0g/mol-

0.135g H * (1mol/1g) = 0.135moles H

Dividing in the moles of N (Lower number of moles) the ratio of atoms is:

N = 0.0300 moles N / 0.0300 moles N = 1

O = 0.0300 moles O / 0.0300 moles N = 1

C = 0.0450 moles C / 0.0300 moles N = 1.5

H = 0.135 moles H / 0.0300 moles N = 4.5

As the empirical formula requires whole numbers, multiplying each ratio twice:

N = 2, O = 2, C = 3 and H = 9

And the empirical formula is:

Answer:

Al + MnO4- + 2H2O → Al(OH)4- + MnO2

Explanation:

First of all, we out down the skeleton equation;

Al + MnO4- → MnO2 + Al(OH)4-

Secondly, we write the oxidation and reduction equation in basic medium;

Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-

Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-

Thirdly, we add the two half reactions together to obtain:

Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-

Lastly, cancel out species that occur on both sides of the reaction equation;

Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O

The simplified equation now becomes;

Al + MnO4- + 2H2O → Al(OH)4- + MnO2

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

Examples

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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.

water is known as the mixture

Answer:

iv. Iron

water is not a mixture

hydrogen is the simplest element

air is pure

Answer:

the mas is .291 g

Explanation:

the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g

Answer:

Methane and chlorine

If a mixture of methane and chlorine is exposed to a flame, it explodes - producing carbon and hydrogen chloride. This is not a very useful reaction! The reaction we are going to explore is a more gentle one between methane and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.

CH4+Cl2+energy→CH3Cl+HCl(3.4.2)

The organic product is chloromethane. One of the hydrogen atoms in the methane has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms.

Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.

I hope this help you!:)

Explanation:

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

Answer:

c

Explanation:

it takes 10,000 years to just reduce down the decay

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.

Answer:

(B). it's metallic bonding

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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