a student combines a solution of nacl(aq) with a solution of agno3(aq) , and a precipitate forms. question which of the following is evidence that ionic bonds formed during the precipitation?

This representation suggests that the element is phosphorus (P) with a mass number of 30, which is incorrect. The correct mass number for phosphorus is approximately 30.97. The incorrect representation for a neutral atom is 36Li

To determine the correct representation for a neutral atom, we need to consider the atomic number (Z) and mass number (A) of the element. The atomic number represents the number of protons in the nucleus, while the mass number represents the sum of protons and neutrons.

Let's analyze the given representations:

36Li:

This representation suggests that the element is lithium (Li) with a mass number of 36, which is incorrect. The correct mass number for lithium is approximately 6.94.

613C:

This representation suggests that the element is carbon (C) with a mass number of 13, which is correct. Carbon has different isotopes, and 13C represents one of its stable isotopes.

3063Cu:

This representation suggests that the element is copper (Cu) with a mass number of 63, which is correct. Copper has different isotopes, and 63Cu represents one of its stable isotopes.

1530P:

This representation suggests that the element is phosphorus (P) with a mass number of 30, which is incorrect. The correct mass number for phosphorus is approximately 30.97.

Therefore, the incorrect representation for a neutral atom is 36Li, as it does not match the known properties of lithium.

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1. The hydronium ion concentration [H₃O⁺] is 4.05×10⁻⁸ M

2. The hydroxide ion concentration [OH⁻] is 1.14×10⁻⁶ M

3. The hydroxide ion concentration [OH⁻] is 1.53×10⁻¹¹ M

4. Solution B is basic

1. How do I determine the value of [H₃O⁺]?

We can obtain the value of [H₃O⁺] as follow:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

Divide both sides by [OH⁻]

[H₃O⁺] = 10¯¹⁴ / [OH⁻]

[H₃O⁺] = 10¯¹⁴ / 2.47×10⁻⁷

[H₃O⁺] = 4.05×10⁻⁸ M

2. How do I determine the value of [OH⁻]?

The value of the hydroxide ion concentration [OH⁻] can be obtained as follow:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

8.81×10⁻⁹ × [OH⁻] = 10¯¹⁴

Divide both side by 8.81×10⁻⁹

[OH⁻] = 10¯¹⁴ / 8.81×10⁻⁹

[OH⁻] = 1.14×10⁻⁶ M

3. How do I determine the value of [OH⁻]?

The value of the hydroxide ion concentration [OH⁻] can be obtained as follow:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

6.55×10⁻⁴ × [OH⁻] = 10¯¹⁴

Divide both side by 6.55×10⁻⁴

[OH⁻] = 10¯¹⁴ / 6.55×10⁻⁴

[OH⁻] = 1.53×10⁻¹¹ M

4.How do I determine which solution is basic?

To k now which solution is basic, we shall determine the pH of each solution. Details below:

pH = –Log [H₃O⁺]

pH = –Log 4.05×10⁻⁸

pH of solution A = 7.4

pH = –Log [H₃O⁺]

pH = –Log 8.81×10⁻⁹

pH of solution B = 8.1

pH = –Log [H₃O⁺]

pH = –Log 6.55×10⁻⁴

pH of solution C = 3.2

Summary:

We know that the pH scale is a scale that gives an understanding of the variation of the acidity / alkalinity of a solution.

The scale ranges from 0 to 14 indicating:

Thus, solution B is basic

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Answer:

I say the second one, im not rlly sure tho

Explanation:

For each color of paper, the part of white light that is reflected depends on the color of the paper. When white light strikes an object, some of the light is absorbed by the object, some of it is transmitted through the object, and some of it is reflected.

The color of the object that we see is the color of the light that is reflected by the object.

For example, when white light strikes blue paper, the blue color of the paper absorbs all the other colors of the spectrum except blue, which is reflected back to our eyes. This is why we see the paper as blue. Similarly, when white light strikes red paper, the red color of the paper absorbs all the other colors except red, which is reflected back to our eyes. This is why we see the paper as red.

In summary, the color of an object is determined by the color of the light that is reflected by the object, and the color of the light that is reflected depends on the color of the object and the colors of the spectrum that are absorbed or transmitted by the object.

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A change in the size of the moon jellies' resource population can change the number of births in the moon jelly population because the big size of the resources can produce more births.

When there is more energy storage molecules present in the moon jellies, they can reproduce more, in more births. Fewer deaths would also lead to the jelly population increasing. The sea turtle population, and the moon jellies consumer population is also decreased.

There must be a change to the birth rate or the death rate in the moon jelly population. Within a population, organisms are born and dying continuously. If the number of births and deaths in a given time interval are equal, then the population size will remain stable.

So we can conclude that a large population of resources will lead to more births.

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We can see here that the catalytic cycle for the hydrogenation of alkenes using Wilkinson's catalyst involves several steps.

Here's an overview of the catalytic cycle, including the necessary details:

i. Chemical structure of the catalyst:

Wilkinson's catalyst, also known as chloridotris(triphenylphosphine)rhodium(I), has the following chemical structure: [RhCl(PPh3)3]

ii. Molecular geometry of the catalyst:

The Wilkinson's catalyst has a trigonal bipyramidal geometry around the rhodium center. The three triphenylphosphine (PPh3) ligands occupy equatorial positions, while the chloride (Cl) ligand occupies an axial position.

iii. Type of reactions involved:

The catalytic cycle involves several reactions, including:

iv. Starting material, reagent, and solvent:

The starting material is an alkene, and the reagent is Wilkinson's catalyst ([RhCl(PPh3)3]). The reaction is typically carried out in a suitable solvent, such as dichloromethane (CH2Cl2) or tetrahydrofuran (THF).

v. Major and minor end-products:

The major end-product of the hydrogenation reaction is the fully saturated alkane, resulting from the addition of hydrogen across the double bond. The minor end-product may include cis- or trans-configured alkanes if the original alkene substrate possesses geometric isomers.

vi. Intermediates:

The intermediates in the catalytic cycle include:

These intermediates play a crucial role in facilitating the hydrogenation reaction and enabling the catalytic cycle to proceed.

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Answer: 0.05 moles

Explanation:

To convert from atoms to moles, you will need Avogadro's number.

Avogadro's number: 6.022×10²³atoms/moles

(3.13×10²²atoms)×(moles/6.022×10²³ atoms)=0.05 moles

The atoms would cancel out and we will be left with our moles.

Answer:

true

Explanation:

Oxoacids are oxygen containing acids .The ionizable hydrogens in sulphuric acid are those bonded to the oxygen atoms. Hence, the statement is correct.

According to Bronsted-Lowry concept acids are substances which can furnish  H+ ions when dissolved in water or on ionization. Oxoacids contains oxygen along with hydrogen atoms.

Some elements have a number of oxoacids mainly for group 15 and 16 elements. Phosphorus and sulphur have 4-5 oxoacids which differ in the number of oxygen and hydrogens.Oxoacids contains the hydroxyl group OH and the naming of oxoacids is based on the number of OH groups.

Sulphuric acid is an oxoacid of Sulphur containing 2 OH groups and two double bonded oxygens. Where the hydrogen atoms can be donated to a base to form its ionizable form .

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The de Broglie wavelength of an electron traveling at 1.59×105m/s is  0.4547 x 10⁻⁹ m.

The elementary electric charge of the electron is a negative one, making it a subatomic particle. Due to their lack of known components or substructure, electrons, which are part of the first generation of the lepton particle family, are typically regarded to be elementary particles.

The length scale at which a particle's wave-like characteristics are significant is indicated by its de Broglie wavelength. The symbol or dB is typically used to indicate the De Broglie wavelength. The de Broglie wavelength for a particle with momentum p is given by dB = hp.

λ = h/mv

Where,

λ = wavelength of electron

m = mass of electron = 9.11e-31 kg

v = speed of electron = 1.59 × 10⁵ m/s

h = constant

Therefore,

λ = (6.626x10⁻³⁴J-s) ÷  [(9.11e-31 kg) (1.59 x 10⁵ m/s)]

λ = 0.4547 x 10⁻⁹ m

Thus, The de Broglie wavelength of an electron traveling at 1.59×105m/s is  0.4547 x 10⁻⁹ m.

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The molar mass of the protein under the conditions that we have in this question can be obtained to be 12065 g/mol.

We know that the osmotic pressure is one the colligative properties of the solution. The colligative properties are the properties that depend on the amount of substance that is present in the solution.

Mass of the protein = 0.135 g

Volume of the solution = 2.00 mL

Temperature of the system = 28 ∘C + 273 = 301 K

Osmotic pressure of the solution = 0.138 atm

We can get the osmotic pressure from;

π = iCRT

π = Omotic pressure

i = Van't Hoff factor

C = concentration of the solution

R = gas constant

T = temperature

Let the molar mass of the protein be M

0.138 = 1 * 0.135/M/0.002 * 0.082 * 301

0.138 =  1 * 0.135/0.002M * 0.082 * 301

0.138 = 3.33/0.002M

0.138 * 0.002M  = 3.33

M = 3.33/0.138 * 0.002

M = 12065 g/mol

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Answer:

4 or 3 atoms; leaning towards 4 more

Explanation:

Answer:

3 atoms

Explanation:

Strong electrolytes : NaCl, HBr The correct options are a and b.

Weak electrolytes : HCH3CO2 The correct option is c.

Non-electrolyte:  CH3OH, H2O The correct options are d and e.

The classification of aqueous solutions as non-electrolytes, strong electrolytes, or weak electrolytes depends on the degree to which they dissociate into ions when dissolved in water.

a) NaCl - Strong electrolyte: It dissociates completely into its ions (Na+ and Cl-) when dissolved in water.

b) HBr - Strong electrolyte: It also dissociates completely into its ions (H+ and Br-) in an aqueous solution.

c) HCH3CO2 (Acetic Acid) - Weak electrolyte: It partially dissociates into its ions (H+ and CH3COO-) in water.

d) CH3OH (Methanol) - Non-electrolyte: It does not dissociate into ions and thus does not conduct electricity in an aqueous solution.

e) H2O (Water) - Non-electrolyte: Although it can act as a weak electrolyte when ionized to a small extent, it is generally considered a non-electrolyte because of its low conductivity.

The correct options for strong electrolytes are a and b, for non-electrolytes are d and e and for weak electrolytes is c.

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Answer: metal turn orange and weaker as it gets oxidised

Explanation:

If a reaction has a negative ΔG and a positive ΔS, the reaction will be spontaneous at all temperatures.

If a reaction is spontaneous at all temperatures, it implies that the reaction will occur without the need for any external intervention, such as the addition of energy. For a reaction to be spontaneous, it must satisfy the criteria of thermodynamic favorability, which is determined by the change in Gibbs free energy (ΔG) associated with the reaction.

The relationship between ΔG, temperature (T), and the equilibrium constant (K) of a reaction is described by the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy and ΔS is the change in entropy.

To ensure spontaneity at all temperatures, two conditions must be met:

ΔG must be negative: A negative ΔG indicates a thermodynamically favorable reaction, meaning the products have a lower Gibbs free energy than the reactants. If ΔG is negative, the reaction will proceed spontaneously in the forward direction.

ΔS must be positive: A positive ΔS signifies an increase in the overall entropy of the system. Higher entropy means more disorder, and spontaneous reactions often involve an increase in randomness. When ΔS is positive, it can compensate for the enthalpic term, ΔH, allowing the reaction to proceed spontaneously.

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its True

Yes, burning of wood is an example of oxidation-reduction reaction.

hope it helps

To answer this question we first look at the reaction:

As we can see the equation is balanced, and for every 1 mol of calcium hydroxide and 2 moles of hydrochloric acid we obtain 2 moles of water.

So now we have to calculate how many moles of calcium hydroxide are 29g, and how many moles of hydrochloric acid are 12.5g.

For this we use these compounds molar mass:

So we calculate the moles of each reactant:

If all the calcium hydroxide reacted then we would need:

But the moles of hydrochloric acid are 0.3429, therefore the limiting reactant is hydrochloric acid.

Now, to answer the question, we calculate the number of moles formed when 0.3429 moles of hydrochloric acid react.

For every mole of HCl that reacts we said that 1 mol of H2O is formed, so the answer to the question is 0.3429 mol of water can be formed.

Approximately 0.157 liters or 157 milliliters of the 4.50 M HCl solution can be made by mixing the given solutions.

To determine the volume of 4.50 M HCl that can be made by mixing the given solutions, we can use the concept of the concentration-volume relationship:

C1V1 = C2V2

Where:

C1 = concentration of the first solution

V1 = volume of the first solution

C2 = concentration of the second solution

V2 = volume of the second solution

Let's assign the variables as follows:

C1 = 5.65 M

V1 = unknown volume (we'll solve for this)

C2 = 3.55 M

V2 = 250.0 mL = 0.250 L (since the volume is given in milliliters)

Now we can plug in the values into the equation and solve for V1:

(5.65 M)(V1) = (3.55 M)(0.250 L)

Dividing both sides of the equation by 5.65 M:

V1 = (3.55 M)(0.250 L) / 5.65 M

V1 ≈ 0.157 L

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As per the given data, 1.6 grams of oxygen reacted in this experiment.

To calculate the amount of oxygen that reacted in the experiment, we need to determine the difference in the mass of X oxide (X,O) formed and the mass of X initially used.

Given:

Mass of X = 4.6 g

Mass of X oxide (X,O) = 6.2 g

To find the amount of oxygen that reacted:

Mass of oxygen = Mass of X oxide - Mass of X

= 6.2 g - 4.6 g

= 1.6 g

Therefore, 1.6 grams of oxygen reacted in this experiment.

Calculate the mass of 1 mole of X:

Given that the mass of X is 4.6 g, we can calculate the molar mass of X by dividing the mass by the number of moles:

Molar mass of X = Mass of X / Number of moles of X

Molar mass of X = 4.6 g / 0.1 mol

Molar mass of X = 46 g/mol

Therefore, the mass of 1 mole of X is 46 grams.

Thus, the answer is 46 grams.

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Answer:

Explanation:

The balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, the number of moles of sodium hydroxide used in the reaction is:

n(NaOH) = c(NaOH) x V(NaOH) = 0.400 mol/L x 0.0200 L = 0.00800 mol

Since one mole of sulfuric acid reacts with two moles of sodium hydroxide, the number of moles of sulfuric acid used in the reaction is twice that of sodium hydroxide:

n(H2SO4) = 2 x n(NaOH) = 2 x 0.00800 mol = 0.0160 mol

The concentration of the sulfuric acid can be calculated by dividing the number of moles by the volume used in the titration:

c(H2SO4) = n(H2SO4) / V(H2SO4) = 0.0160 mol / 0.0250 L = 0.640 M

Therefore, the concentration of the dilute sulfuric acid is 0.640 M.

Answer: The answer is true.

Taking into account the reaction stoichiometry, 0.1129 moles of P₄ is required to react with with 700.0 grams of calcium phosphate.

In first place, the balanced reaction is:

2 Ca₃(PO₄)₂ + 6 SiO₂ + 10 C → 6 CaSiO₃ + 10 CO + P₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: If by reaction stoichiometry 620 grams of Ca₃(PO₄)₂ produces 1 mole of P₄, 70 grams of Ca₃(PO₄)₂ produces how many moles of P₄?

\(moles of P_{4} =\frac{70 grams of Ca_{3}(PO_{4} )_{2}x1 mole of P_{4} }{620 grams of Ca_{3}(PO_{4} )_{2}}\)

moles of P₄= 0.1129 moles

Finally, 0.1129 moles of P₄ is required to react with with 700.0 grams of calcium phosphate.

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Answer:

The formula for calculating pH is pH=−log[H_3O+ ]

pH is the negative logarithm (to base 10) of hydronium ion concentration

The pH Formula can also be expressed as

PH= - log[H+ ]

These substances have dispersed particles that are large enough to scatter light, making the beam visible. Therefore, out of the options provided, a mixture that scatters light is an example of a colloid. Option B)

A colloid is a type of mixture in which particles are dispersed throughout a medium, creating a homogeneous appearance. Unlike solutions, where the particles are completely dissolved, and suspensions, where the particles settle out, colloids have particles that are larger than those in solutions but smaller than those in suspensions. One characteristic of colloids is that they can scatter light due to the size of the particles. This scattering of light is known as the Tyndall effect. Examples of colloids include milk, fog, and aerosol sprays. These substances have dispersed particles that are large enough to scatter light, making the beam visible. Therefore, out of the options provided, a mixture that scatters light is an example of a colloid. Therefore option B) is correct

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Note Complete Question

which is an example of a colloid?

a mixture that settles out,

b mixture that scatters light,

c mixture that is separated by filtration,  

d salt and water mixture?

To graph the function g(x) = f(-x), you can start with the graph of f(x) and then reflect it about the y-axis.

To find the graph of the function g(x) = f(-x), we can start with the graph of the function f(x) and then reflect it about the y-axis.

If the graph of f(x) is symmetric with respect to the y-axis, meaning it is unchanged when reflected, then g(x) = f(-x) will have the same graph as f(x).

However, if the graph of f(x) is not symmetric with respect to the y-axis, then g(x) = f(-x) will be a reflection of f(x) about the y-axis.

In either case, the resulting graph of g(x) = f(-x) will be symmetric with respect to the y-axis.

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The volume of carbon dioxide produced is approximately (d) 11.9 liters.

To determine the amount of carbon dioxide (C\(O_2\)) produced when 37.8 grams of carbon disulfide (C\(S_2\)) reacts with excess oxygen gas (\(O_2\)), we need to use stoichiometry and the given balanced chemical equation:

C\(S_2\)(l) + 3\(O_2\)(g) → C\(O_2\)(g) + 2S\(O_2\)(g)

First, we calculate the number of moles of C\(S_2\) using its molar mass:

Molar mass of (C\(S_2\)) = 12.01 g/mol (C) + 32.07 g/mol (S) × 2 = 76.14 g/mol

Number of moles of (C\(S_2\)) = mass / molar mass = 37.8 g / 76.14 g/mol ≈ 0.496 mol

From the balanced equation, we can see that the stoichiometric ratio between (C\(S_2\)) and C\(O_2\) is 1:1. Therefore, the number of moles of C\(O_2\) produced will also be 0.496 mol.

Now we can use the ideal gas law to calculate the volume of C\(O_2\) at the given temperature and pressure. The ideal gas law equation is:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Converting the temperature from Celsius to Kelvin:

T(K) = 28.85°C + 273.15 = 302 K

Using the ideal gas law:

V = nRT / P = (0.496 mol) × (0.0821 L·atm/mol·K) × (302 K) / (1.02 atm) ≈ 11.9 L

The correct answer is 11.9 liters.

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The reaction is endothermic since the energy level of the products have are higher than that of the reactants.

The activation energy (AE) is the energy difference between the reactants and the transition state.

The change in energy E and the energy difference between the reactants and the products

The data given is as follows:

Reactants: 0 kJ/mol

AE forward 125 kJ/mol

AE reverse: 86 kJ/mol

Products: 39 kJ/mol

The values of ΔE forward and ΔE reverse are as follows:

ΔE forward = (39 - 0) kJ/mol

ΔE forward = +39 kJ/mol

ΔE reverse = (0 - 39) kJ/mol

ΔE reverse =  -39 kJ/mol

3. Given that Ea = 154 kJ/mol and ΔE = 136 kJ/mol

AE reverse = ΔE - AE forward

E = 136 kJ/mol - 154 kJ/mol

E = -18 kJ/mol

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Combustion of 31.68 g of a compound containing only carbon, hydrogen, and oxygen produces 36.67 g CO₂ and 15.01 g H₂O. the empirical formula of the compound is CH₂O.

given that :

mass of CO₂ = 36.67 g

mass of H₂O = 15.01 g

total mass of compound = 31.68 g

moles of CO₂ = 36.67 / 44

                       = 0.833 mol

moles of C = 0.833 mol

moles of  H₂O = 15.01 / 18

                        = 0.833 mol

moles of H = 2(0.833)

                  = 1.667 mol

mass of carbon = 0.833 × 12

                           = 10 g

mass of hydrogen = 1.667 × 1

                               = 1.667 g

mass of oxygen = 31.68 - ( 10 + 1.67 )

                             = 20 g

moles of Oxygen = 20 / 16

                             = 1.25 mol

dividing by the smallest one :

C = 0.833 / 0.833 = 1

H = 2

O = 1

The empirical formula is CH₂O.

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