Which is the electron configuration for bromine? 1s22s22p63s23p64s23d104p5 1s22s22p63s23p64s23d104p6 1s22s22p63s23p44s23d104p5 1s22s22p63s23p64s23d94p5

In order to provide the 300 moles of sodium nitrite required for the reaction using a 0.450 m solution, you would need 666,667 mL of the solution.

To determine how many milliliters of a 0.450 m sodium nitrite solution are needed for a reaction where 300 moles of sodium nitrite are needed, you'll want to use the equation that relates moles, molarity (M), and volume (V). This equation is Moles = Molarity x Volume.

For the given question, we know the amount of moles (300 moles) and the molarity (0.450 m), so we can rearrange the equation and plug in our values to find the volume: Volume (L) = Moles / Molarity = 300 moles / 0.450 M = 666.67 L.

Do remember to convert this volume to milliliters because the question asks for the answer in milliliters. 666.67 L x 1000 mL/L= 666,667 mL.

Therefore, you would need 666,667 mL of a 0.450 m sodium nitrite solution to provide the 300 moles of sodium nitrite required for the reaction.

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To determine the volume of solution needed for the reaction, we can use the formula Volume = Number of moles / Molarity. In this case, the volume of solution needed is approximately 666,666.67 mL.

To determine the volume of solution needed, we need to first calculate the number of moles of sodium nitrite in 300 moles. Since the solution is 0.450 M, this means there are 0.450 moles of sodium nitrite in every 1 liter of solution. Therefore, to calculate the volume of solution needed, we can use the formula:

Volume (in liters) = Number of moles / Molarity

Volume = 300 moles / 0.450 moles

To convert the volume from liters to milliliters, we can multiply the volume by 1000:

Volume (in milliliters) = Volume (in liters) * 1000

Plugging in the values, we get:

Volume (in milliliters) = (300 moles / 0.450 moles) * 1000 = 666666.67 mL

Therefore, approximately 666,666.67 mL of solution are needed for the reaction.

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Answer:

they are transfer from the  towers

Explanation:

Answer:

1) Weak base.

2) Weak acid.

Explanation:

1) Which would have a lower pH, a 0.1 M solution of a strong base or a weak base? Why?

Solution with lower pH means that the solution has a lower basic character.

While solution with higher pH means that the solution has a higher basic character.

When we have a 0.1 M solution of a strong base or a weak base, the weak base will have a lower pH because it is not completely dissociated to produce OH⁻ ions.

While, strong base dissociates completely to produce OH⁻ ions, so it will has higher basic character and thus has a higher pH value.

So, weak base will have the lower pH value.

2) Which would have a higher pH, a 0.1 M solution of a strong acid, or a weak acid? Why?

Solution with lower pH means that the solution has a higher acidic character.

While solution with higher pH means that the solution has a lower acidic character.

When we have a 0.1 M solution of a strong acid or a weak acid, the weak acid will have a higher pH because it is not completely dissociated to produce H⁺ ions, so it has a lower acidic character.

While, strong acid dissociates completely to produce H⁺ ions, so it will has higher acidic character and thus has a lower pH value.

So, weak acid will have the higher pH value.

Answer:

165.726 g.

Explanation:

Cr₂O₃ + 3H₂S → Cr₂S₃ + 3H₂O,

It is clear that 1 mol of Cr₂O₃ and 3 mol of H₂S to produce 1 mol of Cr₂S₃ and 3 mol of H₂O.

no. of moles of Cr₂S₃ = mass/molar mass = (324.8 g)/(200.19 g/mol) = 1.62 mol.

Using cross multiplication:

3 mol of H₂S produces → 1 mol of Cr₂S₃, from stichiometry.

??? mol of H₂S produces → 1.62 mol of Cr₂S₃.

∴ The no. of moles of H₂S are needed = (3 mol)(1.62 mol)/(1 mol) = 4.86 mol.

∴ The "no. of grams" of H₂S are needed = (no. of moles of H₂S)(molar mass of H₂S) = (4.86 mol)(34.1 g/mol) = 165.726 g.

Answer:

Metal A has the highest density.

Explanation:

Density is defined as mass present in unit volume of the substance.

[tex]Density=\frac{Mass}{Volume}[/tex]

Density of metal A

Mass of metal = 122 g

Volume of metal = [tex]10.5 cm^3[/tex]

[tex]D=\frac{122 g}{10.5 cm^3}=11.61 g/cm^3[/tex]

Density of metal B

Mass of metal = 132 g

Volume of metal = [tex]14.2 cm^3[/tex]

[tex]D=\frac{132 g}{14.2 cm^3}=9.29 g/cm^3[/tex]

Density of metal C

Mass of metal = 115 g

Volume of metal = [tex]16.1 cm^3[/tex]

[tex]D=\frac{115 g}{16.1 cm^3}=7.14 g/cm^3[/tex]

Density of metal D

Mass of metal = 126 g

Volume of metal = [tex]12.7 cm^3[/tex]

[tex]D=\frac{126 g}{12.7 cm^3}=9.92 g/cm^3[/tex]

Metal A has the highest density.

Answer: 8

Explanation: Radon has 8 valence electrons. Radon is considered stable with a complete octet of electrons, filling the s and p orbitals.

distillation, magnetic attraction, filtration

Answer: There are C - C  bonds and C - H bonds.

trust me! I took a test with this question and got 100! :)

The pictured compound has C - C  bonds and C - H bonds.

This is defined as an attraction between atoms, ions or molecules which results in the formation of chemical compounds.

The compound is an alkane as the formula is C₄H₁₀(butane) which has only  C - C  bonds and C - H bonds single bonds thereby making option C the most appropriate choice.

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Answer:

1244 students

Explanation:

That would be y = 82*3 + 998

= 1244.

Answer:

The expected student enrollment in 2003 is 1244.

Explanation:

The given diagram it is clear that the graph represents the relationship x and y, where x is number of years since 2000 and y is number of student enrollment.

From the given figure it is clear that the equation of best fit line is

[tex]y=82x+998[/tex]       .... (1)

We need to find the expected student enrollment in 2003.

Year 2003 is 3 years since 2000.

Substitute x=3 in equation (1) to find the expected student enrollment in 2003.

[tex]y=82(3)+998[/tex]

[tex]y=246+998[/tex]

[tex]y=1244[/tex]

Therefore the expected student enrollment in 2003 is 1244.

Answer:

negatively correlated

Answer:

Twice  as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

Answer:

Explanation:

An acid:

A base:

Acids and bases have unique diagnostic characteristics that can be used to identify them. Acids taste sour, turn blue litmus paper red, and react with metals to produce hydrogen gas. Bases taste bitter, turn red litmus paper blue, and feel slippery.

Three diagnostic characteristics of an acid:

Three diagnostic characteristics of a base:

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Answer:

hybrid BB

Dominant Bb

recessive alleles bb

phenotype ratio 3:1

Answer:

hybrid BB

dominant Bb

recessive alleles bb

phenotype ratio 3:1

Explanation:

because phenotype ratio is a type of ratio so it match with the ratio. that's what l think

Answer:

Chemical change is basiclly chemical reaction that creates new substances, sometimes can be hard to observed. The melting of the ice only involves in changing the form of the matter, from solid to liquid form, it doesn't transform the substance which is water into another substance. Therefore, the melting of ice is not a chemical change, but rather a physical one.

Answer:

A new substance is not formed

Explanation:

Its still water just not physically

Answer:

through the water cycle

Explanation:

Answer:

b. CH3OCH3, amine.

Explanation:

It is not an amine It is ethyl ether.

The option that is incorrect pairing is b. CH3OCH3, amine.

CH3OCH3 represents the chemical formula for dimethyl ether, which is a type of ether. Ethers are organic compounds characterized by an oxygen atom linking two alkyl or aryl groups. They are not classified as amines.

Amines, on the other hand, are a different class of organic compounds that contain a nitrogen atom bonded to carbon atoms. They are typically named with the suffix "-amine" and can be primary, secondary, or tertiary depending on the number of alkyl or aryl groups bonded to the nitrogen atom.

Therefore, the pairing of CH3OCH3 with amine is incorrect, as CH3OCH3 is an ether, not an amine.

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Answer:

HNO3 (aq) is an acid

Explanation:

Please mark brainliest and have a great day!

Answer:

See below.

Explanation:

CH4 + 2 O2 → CO2 + 2H2O

One molecule of methane reacts with 2 molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

It is also true if we replace the word molecule by mole.

Answer : The correct option is,

One molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

One mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and 2 moles of water.

Explanation :

The given balanced chemical reaction is,

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

The given chemical reaction is a balanced reaction.

By the stoichiometry we can say that,

One molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

or,

One mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and 2 moles of water.

Answer:

Pastoral farming is best described as a method of raising livestock and moving herds. (a.)

Answer:

B) oxidized

Explanation:

When glucose loses a hydrogen atom, it is losing 1 proton and 1 electron. This loss of an electron means it is oxidized.

The glucose molecule becomes oxidized when it loses a hydrogen atom in an oxidation-reduction reaction. This reflects the loss of electrons, which is characteristic of oxidation, while gaining electrons, or hydrogen atoms, indicates reduction.(Option b)

When a glucose molecule loses a hydrogen atom as the result of an oxidation-reduction reaction, the molecule becomes oxidized. This loss of hydrogen represents the molecule giving up electrons, which characterizes the process of oxidation. Conversely, when a molecule gains hydrogen atoms, it is said to be reduced.

In the context of glucose metabolism, an oxidizing agent would be a molecule like NAD+ or FAD, which accepts the electrons that are lost when glucose is oxidized. Hence, the carbohydrate is oxidized, liberating energy that the body can use.

Remember that in the process of cellular respiration, glucose is oxidized into carbon dioxide (CO2), and the oxygen is reduced while forming water (H2O). Glucose losing hydrogens means it is becoming oxidized, and the gaining of hydrogen by oxygen to form water indicates that oxygen is being reduced.

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Answer:

pH = 6.81.

Explanation:

H₂O(l) ⇄ H⁺(aq) + OH⁻(aq),

Kw = [H⁺][OH⁻] = 2.4 x 10⁻¹⁴.  

in pure water and neutral aqueous solution, [H⁺] = [OH⁻]  

So, Kw = [H⁺]²

∴ 2.4 x 10⁻¹⁴ = [H⁺]²

∴ [H⁺] = 1.55 x 10⁻⁷ M.

∵ pH = - log [H⁺]  

pH = - log (1.55  x 10⁻⁷) = 6.81.

Answer:

[tex]\boxed{\text{(a) 209 mL; (b) } 6.09 \times 10^{23}}[/tex]

Explanation:

(a) Gas produced at cathode.

(i). Identity

The only species known to be present are Cu, H⁺, and H₂O.

Only the H⁺ and H₂O can be reduced.

The corresponding reduction half reactions are:

(1) 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻;     E° = -0.8277 V

(2) 2H⁺ +2e⁻ ⇌ H₂;                     E° =  0.0000 V

Two important points to remember when using a table of standard reduction potentials:

H⁺ is below H₂O, so H⁺ is reduced to H₂.

The cathode reaction is 2H⁺ +2e⁻ ⇌ H₂, and the gas produced at the cathode is hydrogen.

(ii) Volume

a. Anode reaction

The only species that can be oxidized are Cu and H₂O.

The corresponding half reactions  are:

(3) Cu²⁺ + 2e⁻ ⇌ Cu;                E° =  0.3419 V

(4) O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O     E° =   1.229   V

Cu is above H₂O, so Cu is more easily oxidized.

The anode reaction is Cu ⇌ Cu²⁺ + 2e⁻.

b. Overall reaction:

Cu           ⇌ Cu²⁺ + 2e⁻

2H⁺ +2e⁻ ⇌ H₂                    

Cu + 2H⁺ ⇌ Cu²⁺ + H₂

c. Moles of Cu lost

[tex]n_{\text{Cu}} = \text{0.584 g } \times \dfrac{\text{1 mol}}{\text{63.55 g}} = 9.190 \times 10^{-3}\text{ mol Cu}[/tex]

d. Moles of H₂ formed

[tex]n_{\text{H}_{2}}} = 9.190 \times 10^{-3}\text{ mol Cu} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol Cu}} =9.190 \times 10^{-3}\text{ mol H}_{2}[/tex]

e. Volume of H₂ formed

Volume of 1 mol at STP (0 °C and  1 bar) = 22.71 mL

[tex]V = 9.190 \times 10^{-3}\text{ mol}\times \dfrac{\text{22.71 L}}{\text{1 mol}}  = \text{0.209 L} = \boxed{\textbf{209 mL}}[/tex]

(b) Avogadro's number

(i) Moles of electrons transferred

[tex]\text{Moles of electrons} = 9.190 \times 10^{-3}\text{ mol Cu}\times \dfrac{\text{2 mol electrons}}{\text{1 mol Cu}}\\\\\\= \text{0.018 38 mol electrons}[/tex]

(ii) Number of coulombs

Q  = It  

Q = \text{1.18 C/s} \times 1.52 \times 10^{3} \text{ s} = 1794 C

(iii). Number of electrons

[tex]n = \text{ 1794 C} \times \dfrac{\text{1 electron}}{1.6022 \times 10^{-19} \text{ C}} = 1.119 \times 10^{22} \text{ electrons}[/tex]

(iv) Avogadro's number

[tex]N_{\text{A}} = \dfrac{1.119 \times 10^{22} \text{ electrons}}{\text{0.018 38 mol}} = \boxed{6.09 \times 10^{23} \textbf{ electrons/mol}}[/tex]

Answer:

(4) 2.0 M CaCl₂(aq).

Explanation:

ΔTb = i.Kb.m,

where, ΔTb is the elevation in boiling point.

i is the van 't Hoff factor.

Kb is the molal elevation constant of water.

m is the molality of the solution.

(1) 1.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(1.0 m) = 2(Kb).

(2) 2.0 M KCl(aq):

i for KCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (2)(Kb)(2.0 m) = 4(Kb).

(3) 1.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 1.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(1.0 m) = 3(Kb).

(4) 2.0 M CaCl₂(aq):

i for CaCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

suppose molarity = molality, m = 2.0 m,

∴ ΔTb for (1.0 M KCl) = i.Kb.m = (3)(Kb)(2.0 m) = 6(Kb).

The aqueous solution with the highest boiling point at standard pressure is the 2.0 M CaCl2 solution because it has the highest concentration and dissociates into the most ions.

The boiling point of an aqueous solution is influenced by the concentration of solutes due to the colligative property known as boiling point elevation. According to the principle of boiling point elevation, each mole of solute particles will raise the boiling point of 1 liter of water by 0.51°C. Taking this into account, the boiling point of a solution is also affected by the number of particles into which each formula unit dissociates in solution. For example, CaCl2 dissociates into three ions (one calcium ion and two chloride ions).

Given the options, the aqueous solution with the highest boiling point at standard pressure is the 2.0 M CaCl2 solution (Option 4). This is because it has the highest concentration of particles in solution: for every mole of CaCl2 there are three moles of ions due to dissociation, and its molarity is the highest among the given solutions.

Answer:

a.) excess reactant aqueous copper chloride.

b.) excess reactant aluminum.

d.) product solid copper.

f.) product aqueous aluminum chloride.

g.) water.

Explanation:

just did the lab, here's the answers :)

Excess reactant aluminum, oxygen, product solid copper, product aqueous aluminum chloride, water could remain at the end of the reaction.

Explanation:

When performing the reaction between aluminum and copper chloride, there are several substances that could remain at the end of the reaction:

Excess reactant aluminum: If aluminum is present in excess, then it would remain at the end of the reaction.

Oxygen: If oxygen is introduced into the reaction, it could remain as a result of incomplete combustion.

Product solid copper: Copper is formed as a product of the reaction between aluminum and copper chloride.

Product aqueous aluminum chloride: Aluminum chloride is also formed as a product of the reaction.

Water: Water can also be present as a product depending on the conditions of the reaction.

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Answer: The correct answer is hydrogen.

Explanation:

pH is defined as the negative logarithm of hydrogen or hydronium ion concentration that are present in a solution.

The equation representing pH of the solution follows:

[tex]pH=-\log[H^+][/tex]

pH range of the solution varies from 0 to 14.

If the pH of the solution is from 0 to 6.9, the solution is said to be acidic.

If the pH of the solution is from 7.1 to 14, the solution is said to be basic.

If the pH of the solution is 7, the solution is said to be neutral.

Thus, different concentration of hydrogen ions will give us different values of pH.

Hence, the correct answer is hydrogen.

The different pH values for the test solutions result from the distinct concentrations of hydrogen (H+) or hydroxide (OH-) ions present in each solution. Acidic solutions have more H+ ions and a lower pH, while alkaline solutions have more OH- ions and a higher pH.

The different pH values for the test solutions are caused by the varying concentrations of their hydrogen ions (H+) or hydroxide ions (OH-). The pH value is essentially a measure of the hydrogen ion concentration in the solution. Acidic solutions have a higher concentration of H+ ions and therefore a lower pH value (below 7), while basic or alkaline solutions have a higher concentration of hydroxide ions, resulting in pH values above 7. For instance, if one solution has more hydroxide ions than another, it will have a higher pH value, indicating it is more basic.

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Answer:

they all contain the same number of molecules at constant temperature and pressure

Answer:

Option D is correct! - They have different masses.

Explanation:

Answer: The total reduction potential of the cell is -3.89 V.

Explanation:

We are given:

[tex]Li^++e^-\rightarrow Li[/tex]

The standard reduction potential for this is -3.04 V

[tex]Hg\rightarrow Hg^{2+}+2e^-[/tex]

The standard reduction potential for this is -0.85 V

The cell formed by these half reactions is: [tex]Hg/Hg^{2+}||Li^+/Li[/tex]

The cell potential, [tex]E^o_{cell}=E^o_{oxidation}+E^o_{reduction}[/tex]

[tex]E^o_{cell}=[-0.85+(-3.04)]=-3.89V[/tex]

Hence, the total reduction potential of the cell is -3.89 V.

Answer:

It is believed that the stability of an isotope is based on the ratio of neutrons to protons.

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Answer: Electrons and protons

An atom of lithium-7 has an equal number of electrons and protons. Therefore, the correct option is option 2.

Lithium is a chemical substance with the atomic number three as well as the symbol Li. The material is a delicate alkali metal in white-silver. Under normal conditions, it is both the lowest population metal and the least populated inorganic element. Because lithium is an extremely reactive as well as flammable alkali metal, it must be stored in a vacuum.

When cut, it has a glossy luster, but wet air quickly corrodes it, turning it into an olive drab silvery gray, eventually merely a black patina. Only compounds, particularly pegmatitic minerals, which has also historically been the primary source of lithium, may exist naturally. An atom of lithium-7 has an equal number of electrons and protons.

Therefore, the correct option is option 2.

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Answer:

1.152 M.

Explanation:

∴ (MV) before dilution = (MV) after dilution

before dilution: M = 6.0 M, V = 48.0 mL.

after dilution: M = ??? M, V = 0.25 L = 250.0 mL.

∴ M after dilution = (MV) before dilution/(V) after dilution = (6.0 M)(48.0 mL)/(250.0 mL) = 1.152 M.

To calculate the molarity of the diluted solution, use the formula M1V1 = M2V2 and plug in the values to find the final molarity.

To calculate the molarity of the diluted solution, we can use the formula:

M1V1 = M2V2

Where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Plugging in the values:

(6.00 M)(48.0 mL) = (M2)(0.250 L)

and solving for M2, we find:

M2 = (6.00 M)(48.0 mL) / (0.250 L) = 1152 M

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