If d represents the density of a gas and k is a constant. The relationship between the rate of diffusion r, and d is ____?

With an Ag concentration of 0.020 M and a Pb2+ concentration of

0.0010 M, Only PbSO4(s) will precipitate,

Option C is correct

Generally, the equation for the Chemical Reaction  is mathematically given as

Ag2SO4--->2Ag++SO4^{-2}

Where

PbSO4--->Pb2+ +SO4^{-2}

Therefore

Ag2SO4=[0.02^2[0.001]]

Ag2SO4=4*10^{-7}

and

PbSO4=[0.001^2[0.001]]

PbSO4=4*10^{-6}

In conclusion,Since PbSO4>>PAg2SO4, Only PbSO4(s) will precipitate.

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Answer:

°C = °K - 273.

Explanation:

the 1st and 2d rows belong to F°; the last one is wrong; correct answer is °C=°K-273 only.

For example, 0°C=273°K; 10°C=283°K; 100°C=373°K.

Answer:

°C=K+273

Explanation:

To convert to °C add 273

The net ionic equation for the reaction between potassium sulfide and magnesium iodide is S2- + Mg2+ -> MgS, as the potassium and iodide ions are spectator ions and do not participate in the reaction.

To determine the net ionic equation for the reaction between potassium sulfide (K2S) and magnesium iodide (MgI2), we first need to identify the ions present in each compound and then determine the products formed when they react.

Potassium sulfide (K2S) dissociates into two potassium ions (K+) and one sulfide ion (S2-):

K2S -> 2K+ + S2-

Magnesium iodide (MgI2) dissociates into one magnesium ion (Mg2+) and two iodide ions (I-):

MgI2 -> Mg2+ + 2I-

Now, we need to determine the possible products when these ions combine. Since potassium (K+) has a +1 charge and iodide (I-) has a -1 charge, they can combine to form potassium iodide (KI):

K+ + I- -> KI

Similarly, magnesium (Mg2+) and sulfide (S2-) can combine to form magnesium sulfide (MgS):

Mg2+ + S2- -> MgS

Now, we can write the complete ionic equation by representing all the ions present before and after the reaction:

2K+ + S2- + Mg2+ + 2I- -> 2KI + MgS

To obtain the net ionic equation, we remove the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the actual reaction. In this case, the spectator ions are the potassium ions (K+) and the iodide ions (I-).

Thus, the net ionic equation for the reaction between potassium sulfide and magnesium iodide is:

S2- + Mg2+ -> MgS

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The percentage of the stannous fluoride present is  1.28 %.

We know that the stoichiometry of a reaction can be used to obtain the relationship that exists between the reactants and the products of a reaction. In this case, the amount of the stannous fluoride that was involved at the beginning can be obtained by the use of the stoichiometry of the balanced chemical reaction.

Number of moles of the stannous phosphate = 0.075-g/546.07 g/mol

= 1.37 * 10^-4 moles

We have from the reaction equation that;

3 moles of  stannous fluoride produces 1 mole of stannous phosphate

x moles of stannous fluoride produces 1.37 * 10^-4 moles of stannous phosphate

x = 3 *  1.37 * 10^-4 /1

x = 4.11 * 10^-4 moles

Molar mass of  stannous fluoride = 156.69 g/mol

Mass of the stannous fluoride = 156.69 g/mol *  4.11 * 10^-4 moles

= 0.064 g

Percentage of stannous fluoride = 0.064 g/ 5.000-g * 100/1

= 1.28 %

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Exactly 149.6J will raise the temperature of 10.0g of a metal from 25.0C. The specific heat capacity of the metal is 5.984 J/g°C.

The heat capacity of a sample of a substance divided by the mass of the sample yields the specific heat capacity (symbol c), also known as massic heat capacity. Informally, it is the quantity of heat that must be added to one unit of a substance's mass in order to raise its temperature by one unit. The specific heat capacity unit in the SI is the joule per kelvin per kilogram, or Jkg⁻¹K⁻¹. For instance, the specific heat capacity of water is 4184 J kg⁻¹K⁻¹, or the amount of energy needed to raise 1 kilogram of water by 1 K.

The specific heat capacity of the metal can be calculated using the equation Q = m × c ×ΔT.

Q = 149.6J

m = 10.0g

ΔT = (final Temperature - initial Temperature) = (25°C - 0°C) = 25°C

Plugging these values into the equation, we get:

149.6J = 10.0g × c ×25°C

Solving for c, we get:

c =  \(\frac{149.6J}{(10.0g *25C)}\)

c = 5.984 J/g°C

Therefore, the specific heat capacity of the metal is 5.984 J/g°C.

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Answer:

What are the following substances?

Explanation:

Answer:

since,hydrogen bonding requires a strongly electronegative atom bonded to a hydrogen atom,only HF willexhibit hydrogen bonding,as afluorine is sufficiently partially positive hydrogen that can form bondwith a lone pair an oxygen in water .

Hope this help u

\(\\ \tt\hookrightarrow Speed=\dfrac{Distance}{Time}\)

\(\\ \tt\hookrightarrow Speed=\dfrac{6}{2}\)

\(\\ \tt\hookrightarrow Speed=3m/s\)

Answer:

Given, Distance travelled(d)=6m

Time taken(t)=2sec

Speed(s)=?

We know that,

\( \purple {speed(s) = \frac{distance \: travelled(d)}{time \: taken(t)}} \)

\( \purple{speed(s) = \frac{6m}{2sec} }\)

\( \purple{speed(s) = 3 m /s}\)

The amount of heat energy produced when 1 g of hydrogen and 2 g of nitrogen are reacted, is -6.61 KJ

First, we shall obtain the limiting reactant. Details below:

3H₂ + N₂ -> 2NH₃

From the balanced equation above,

28 g of N₂ reacted with 6 g of H₂

Therefore,

2 g of N₂ will react with = (2 × 6) / 28 = 0.43 g of H₂

We can see that only 0.43 g of H₂ is needed in the reaction.

Thus, the limiting reactant is N₂

Finally, we the amount of heat energy produced. Details below:

3H₂ + N₂ -> 2NH₃ ΔH = -92.6 KJ

From the balanced equation above,

When 28 grams of N₂ reacted, -92.6 KJ of heat energy were produced.

Therefore,

When 2 grams of N₂ will react to produce = (2 × -92.6) / 28 = -6.61 KJ

Thus the heat energy produced from the reaction is -6.61 KJ

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Answer:

283.5 of radioisotopes. Each mL has 105 micrograms, so multiply 105 x 2.70.

Answer:

.

Explanation:

Covalent bonds are formed when two or more atoms (often...) share electrons. Better (more stable) than larger formal charges are smaller ones.

The electron geometry is the foundation for hybridization. Observe the VSEPR Table to ascertain the hydridization of a molecule's core atom. How do sp3 orbitals come into being? The combination of one s and three p orbitals results in four sp3 orbitals.

When a molecule's geometry shows how bond pairs and lone pairs are arranged around the central atom as well as the coordination number of the molecule. The lone pairs around the central atom.

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Answer:

an increase in temperature

Explanation:

The chemical reaction that is produced by modelage masks causes: an increase in temperature

Answer:

MM = 225.11 g/mol

Explanation:

In this case, let's analyze the data.

We have a 15.2 mg of an unknown electrolyte in 150 mL of solution. The osmotic pressure is 8.44 Torr, and we want the molecular mass of the unknown.

The osmotic pressure can be calculated using the following expression:

π = CRT   (1)    

π: osmotic pressure (8.44 Torr * 1 atm / 760 Torr = 0.011 atm)

C: Concentration of the unknown in the solution

R: universal constant of gases (0.082 L atm / K mol)

T: temperature in Kelvin (25 + 273 = 298 K)

From this expression, we can either solve for C, and then use another expression to calculate the molecular mass, or we can just replace the expressions in the above formula, to get the direct molecular mass. In this case, we'll follow the second method.

Concentration or molarity of a substance can be calculated using:

C = moles / V  (2)

And moles can be calculated using this expression:

moles = m / MM   (3)

Replacing (3) in (2), and then in (1) we have:

C = m / MM * V

π = m * RT / MM * V   (4)

and now, we can solve for MM:

MM = mRT / π V  (5)

Now, we just need to replace the given data into the above expression to get the value of the molecular mass:

MM = (15.2 / 1000) * 0.082 * 298 / 0.011 * 0.150

Hope this helps

\(K_{sp}\) of the compound is found to be  5.04 ×\(10^{-10}\).

Solubility :Solubility can be define as the amount of a substance that dissolves or mixes in a given amount of solvent at specific conditions.

Solubility equilibrium

Ksp = \([A^{+} ]^{a}\) \([B^{-} ]^{b}\)

Ksp = solubility product constant

A+ = cation in an aquious solution

B- = anion in an aqueous solution

a, b = relative concentrations of a and b

Given,

Solubility = s = 3.60 × \(10^{-2}\) g/L

molar mass = 288 g/ mol

∴ s= 3.60 × \(10^{-2}\) g/L ÷ 288 g/ mol = 1.25 ×\(10^{-4}\) mol/ L

Reaction:

MX3 ⇄ M + 3X

           s       3s

\(K_{sp}\) =[ \(M^{+3}\)] [ \(X^{-1}\)\(]^{3}\) = solubility product

∴ \(K_{sp}\) =\([s]^{} [3s]^{3}\)

∴ \(K_{sp}\) = 3 \(s^{4}\)

∴ \(K_{sp}\) = 3 × (3.60 × \(10^{-2}\) \()^{4}\)

∴ \(K_{sp}\) = 503.8848 ×\(10^{-8}\)  = 5.04 ×\(10^{-10}\)

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Answe

I hope this is useful! :)

Explanation:

We use the equation:

Mass = Number of moles * Molecular or atomic mass

Since the number of moles is 1 ("one mole of carbon-12 atoms") and the atomic mass of carbon-12 atoms is 12, hence the mass would be 12 grams. Note that the unit grams (g) is used here, as it is the SI unit for mass measurement.

Number of particles in 1 mole = 6.023 × 1023

Mass of 1 electron = 9.10938 × 10-28 gm

Mass of 1 mole of electrons = 6.023 × 1023 × 9.10938 × 10-28  = 5.486  × 10 -4

Mass of 1 proton = 1.67262 × 10 -24 gm

Mass of 1 mole of protons = 6.023 × 1023 × 1.67262 × 10 -24 gm = 1.007 gm

Mass of 1 neutron = 1.67493 × 10-24  gm

Mass of 1 mole of neutron = 1.67493 × 10-24 × 6.023 × 1023   = 1.008 gm

The mass of one mole of carbon -12 atoms is less than the sum of masses of the particles it contains.

One mole of substance contains 6.022×10²³atoms.Therefore mass of one mole carbon-12 can be calculated as,

                 mass=number of moles ×molar mass

                        =1×12=12 g

Mass of one mole of carbon -12 is 12 g.

Carbon -12 has 6 electrons,  6 protons and  6 neutrons.Mass of each sub atomic particle is as follows,

Mass of electron =0.0005 u

Mass of proton=1.0072 u

Mass of neutron=1.0087 u

As carbon -12 has 6 protons,electrons and neutrons each the mass of each of the subatomic particle is multiplied by 6.

Sum of masses of  particles =6×0.0005+6×1.0072+6×1.0087=12.0984 u

where in,mass of each particle is multiplied by 6 and then added.

Thus, the mass of one mole of carbon -12 is less than the sum of masses of it's particles .

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The pressure of the gas when the temperature changes from -75.0°C to 1000.0°C will be approximately 9.51 atm.

Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.

It is expressed as;

P₁/T₁ = P₂/T₂

Given that:

We substitute our values into the expression above.

P₁/T₁ = P₂/T₂

P₁T₂ = P₂T₁

\(P_2 = \frac{P_1T_2}{T_1}\\ \\P_2 = \frac{1.480\ *\ 1273.15 }{198.15} \\\\P_2 = 9.51 \ atm\)

Therefore, the final pressure is 9.51 atm.

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Answer:

Tcs foods are foods that pose a greater risk of causing foodborne illness if not prepared.

Non Tcs foods on the other hand, are foods that are less likely to support the growth of bacteria and have a lower risk of causing foodborne illness.

Answer:

0.15 g of CuSO4.

Explanation:

What is given?

Concentration of CuSO4 = 0.00095 mol/L.

Liters of solution = 1 L.

RAM (Cu) = 63.

RAM (S) = 32.

RAM (O) = 16.

Step-by-step solution:

Let's see the formula of molarity (concentration of solution):

And we want to find the moles of solute to calculate the weight of copper sulfate. The statement is telling us that we're going to use 1 liter of solution, so let's solve for 'moles of solute' and replace the given data (molarity of CuSO4 solution = concentration of CuSO4 = 0.00095 mol/L and liters of solution = 1L):

Now, we have to convert 0.00095 moles of CuSO4 to mass and here we're going to use the RAM (relative atomic mass)

First, let's calculate the RAM of CuSO4 by doing an algebraic sum with the given RAMs. You can see that we have 1 Cu, 1 S, and 4 O, so the calculation will look like this:

The RAM of CuSO4 is 159. The value of RAM is equivalent to molar mass, but the difference is that the molar mass has the units of g/mol, so, in this case, the molar mass of CuSO4 is 159 g/mol. So now, the conversion from moles to grams will be:

The answer is that they have to weigh out 0.15 g of CuSO4.

Answer:

Explanation:

Gravity pulls the balls down the ramp, and the force of gravity is bigger on larger-mass objects. The extra force on the bigger ball means that it has more energy when it gets to the bottom of the ramp and consequently travels more before stopping.

First, we need to write out the balanced chemical equation for the reaction: 4 KO2 + 2 CO2 → 2 K2CO3 + 3 O2

From the equation, we can see that 4 moles of KO2 react with 2 moles of CO2 to produce 3 moles of O2. Therefore, we need to convert the given masses of KO2 and CO2 into moles:

moles of KO2 = 2.61 g / molar mass of KO2 = 2.61 g / 71.10 g/mol = 0.0367 mol
moles of CO2 = 4.46 g / molar mass of CO2 = 4.46 g / 44.01 g/mol = 0.1013 mol

Next, we need to determine the limiting reagent (the reactant that will be completely consumed in the reaction) by comparing the mole ratios of KO2 and CO2 in the balanced equation. The ratio of moles of KO2 to moles of CO2 is:
0.0367 mol KO2 / 4 mol KO2 per 2 mol CO2 = 0.0184 mol CO2

Since this ratio is less than the actual number of moles of CO2 we have (0.1013 mol), CO2 is in excess and KO2 is the limiting reagent.

Using the mole ratio from the balanced equation, we can calculate the number of moles of O2 produced:

moles of O2 = 3 mol O2 per 4 mol KO2 × 0.0367 mol KO2 = 0.0275 mol O2

Finally, we can convert the moles of O2 to grams:

mass of O2 = moles of O2 × molar mass of O2 = 0.0275 mol × 32.00 g/mol = 0.88 g
Therefore, 2.61 g of KO2 and 4.46 g of CO2 would produce 0.88 g of O2.

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In an acetic acid solution:

1. To make a buffer with pH = 5.00, have a ratio of

\(\frac{[A-]}{[HA]} = 10^{-5.50}\) = 0.316.

The volume of sodium hydroxide needed:

V(NaOH) = (0.316 M - 0.200 M) / 4.50 M = 0.0316 L = 31.6 mL

Therefore, 31.6 mL of 4.50 M sodium hydroxide must be added to 250.0 mL of 0.200 M acetic acid solution to make a buffer with pH = 5.00.

2. The pH of the buffer is calculated as follows:

pH = pKa + log(\(\frac{[A-]}{[HA]}\))

= 4.76 + log(0.2/0.1)

= 4.86

Therefore, the pH of the buffer is 4.86.

3. The mass of salt that must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87:

\(\frac{[A-]}{[HA]} = 10^{-4.87}\) = 0.0114

Therefore, the mass of acetate that must be dissolved:

Mass of acetate = (0.0114 mol dm³)(0.25 dm³) = 0.00285 g

Therefore, 0.00285 g of sodium propanoate must be dissolved in 0.25 dm³ of 1 mol dm³ propanoic acid to give a buffer of pH 4.87.

4. The pH of the buffer is calculated as follows:

pH = pKa + log(\(\frac{[A-]}{[HA]}\))

= 4.74 + log(0.1/0.1)

= 4.74

Therefore, the pH of the buffer is 4.74.

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Option B, where [OH-] is 1.0 x 10-13 mol dm-³3, is the only one that can be considered basic. Therefore, Option B is the correct answer.

To determine whether a solution is basic or acidic at 25 °C, we can compare the concentration of hydroxide ions ([OH-]) with the concentration of hydronium ions ([\(H_3O\)+]). In a neutral solution, the concentrations of [\(H_3O\)+] and [OH-] are equal, resulting in a pH of 7.

Option A states that the concentration of [\(H_3O\)+] is 1.0 x 10-3 mol dm-3. Since [\(H_3O\)+] represents the concentration of hydronium ions, this solution would be acidic because the concentration of [\(H_3O\)+] is higher than [OH-], indicating an excess of hydronium ions.

Option B states that the concentration of [OH-] is 1.0 x 10-13 mol dm-³3. In this case, [OH-] is higher than [\(H_3O\)+], indicating an excess of hydroxide ions. Therefore, this solution would be considered basic.

Option C states that the solution has a pH of 4.00. A pH of 4.00 is below the neutral pH of 7, indicating an excess of hydronium ions and an acidic solution. Therefore, this option does not represent a basic solution.

Option D states that the concentration of [\(H_3O\)+] is 1.0 x 10-13 mol dm-3. Similar to Option A, this concentration of [\(H_3O\)+] indicates an acidic solution, not a basic one.

Option B

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Answer:

D is the correct answer

Explanation:

In order for a solution to be basic at 25 C, the H+ concentration has to be less than the OH- concentration, and given that H+ times OH- is 10^-14, we deduce that H+ must be less than 10^-7 for the solution to be acidic. Thus, A can be eliminated, and so can C. With B, we calculate an H+ concentration of 0.1M, which also fails to be less than 10^-7
Thus, D is the correct answer and we can verify that as H+ is less than 10^-7.



Note: I do not know why my previous answer was deleted for "being incorrect", and i'm not sure how the incorrect answer was "expert verified", but I am as certain that D is the correct answer as i am sure of 3*(4+5-1) being equal to 24.

The observation that compound A releases H2 gas while compound B does not when treated with Na metal can be explained by considering the structural differences between the two compounds and their ability to undergo specific reactions.

Compound A and compound B both have the molecular formula C₂H₆O, which indicates that they both contain two carbon atoms, six hydrogen atoms, and one oxygen atom. However, the difference lies in the arrangement of these atoms within the molecules. One possible explanation for the observed difference is that compound A is an alcohol, specifically ethanol (CH₃CH₂OH), while compound B is an ether, such as dimethyl ether (CH₃OCH₃). The presence of the hydroxyl group (-OH) in ethanol enables it to undergo a reaction with sodium metal, known as the metal-acid reaction. In this reaction, the metal displaces the hydrogen from the hydroxyl group, forming sodium ethoxide (CH₃CH₂ONa) and releasing hydrogen gas (H₂). On the other hand, ethers like dimethyl ether lack the hydroxyl group and therefore cannot undergo the metal-acid reaction. Consequently, when compound B is treated with sodium metal, no hydrogen gas is released. The ability of compound A to release hydrogen gas while compound B does not when treated with sodium metal can be attributed to the presence of a hydroxyl group in compound A (ethanol), enabling it to undergo a metal-acid reaction, whereas compound B (dimethyl ether) lacks the necessary functional group and thus does not undergo this reaction.

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The chloride ion (Cl-) has a single negative charge and a full octet of electrons in its outermost shell. It has a tetrahedral shape, with three equatorial lone pairs and one axial bond to a hydrogen or other positively charged ion.

The conjugate acid of chloride ion is HCl (hydrogen chloride), which is formed when the chloride ion accepts a proton (H+) from an acid. The resulting molecule has a positive charge and a linear shape, with a single bond between the hydrogen and chlorine atoms.

The conjugate acid of chloride ion, HCl, is a strong acid that readily donates a proton (H+) to a base to form the chloride ion. In water, HCl dissociates completely to form H+ and Cl- ions. The hydrogen ion (H+) has a positive charge and no electrons, while the chloride ion (Cl-) retains its original tetrahedral shape and three lone pairs of electrons.

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By using the ideal gas law and Avogadro's number, we find that there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

To determine the number of molecules of CO2 in 2.5 L at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's number.

Avogadro's number (N_A) is a fundamental constant representing the number of particles (atoms, molecules, ions) in one mole of substance. Its value is approximately 6.022 × 10^23 particles/mol.

STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (1 atm).

First, we need to convert the volume from liters to moles of CO2. To do this, we use the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Since we have STP conditions, we can substitute the values:

(1 atm) × (2.5 L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K).

Simplifying the equation:

2.5 = n × 22.4149.

Solving for n (the number of moles):

n = 2.5 / 22.4149 ≈ 0.1116 moles.

Next, we can calculate the number of molecules using Avogadro's number:

Number of molecules = n × N_A.

Number of molecules = 0.1116 moles × (6.022 × 10^23 particles/mol).

Number of molecules ≈ 6.72 × 10^22 molecules.

Therefore, there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

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When the second cation first starts to precipitate, 80.8% of Ca²⁺ will still be in solution.

A cation is an ion with a positive charge. It is formed when an atom loses one or more of its electrons, resulting in a net positive charge. Cations are attracted to anions (ions with a negative charge) due to their opposite charges. Cations are found in many different substances, including acids, bases, and salts.

Ca₃(PO₄)₂ will be the first species that separates out of solution when solid Na₃PO₄ is introduced to the mixture. This is due to Ca3(PO4)2 having a substantially lower solubility than Ag₃PO₄ and Na₃PO₄.

The proportion of the first cation (Ca²⁺ ) still in solution when the second cation (Ag⁺) is just beginning to precipitate will depend on the starting concentrations of the two cations. In this instance, the starting concentrations of Ca²⁺  and Ag⁺ are 0.0400 M and 0.0990 M, respectively. Therefore, 80.8% of Ca²⁺  will still be in solution when its second cation first begins to precipitate.

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