Calculate the hydronium ion concentration, [H3O+] , for a solution with a pH of 7.91 .

Explanation:

the answer is in the image above

Answer:

Assuming the noise remains constant, the total noise for the longer path length, 65 scans should be 0.0005 x 65 = 0.0325 absorbance units. The new reading should be 0.01 absorbance units. The signal to noise ratio (SNR) of the new measurement will be 0.01/0.0325 = 3.08.

Signal averaging decreases the magnitude of the residual noise and increases the SNR. The total noise decreases by the square root of the number of scans, in this case √65 = 8, so for 65 scans, the noise level is 8 times lower than for a single scan.

From the equation 2PbS + 3O2 ⟶ 2Pb + 2SO3 we can produced 8.12 g of lead  if 2.54 g of PbS is burned with 1.88 g of O2.

To find the mass of lead produced, we need to find the limiting reactant (i.e. the reactant that is consumed first).

We can find the limiting reactant by calculating the number of moles of each reactant and then using the mole ratio from the balanced equation.

Number of moles of PbS = mass / molar mass = 2.54 / 239.27 = 0.0106 mol

Number of moles of O2 = mass / molar mass = 1.88 / 32 = 0.0588 mol

From the balanced equation, the mole ratio of PbS to O2 is 2:3.

Therefore, for every 2 moles of PbS, we need 3 moles of O2.

We can use this information to calculate how many moles of O2 are needed for 0.0106 moles of PbS.0.0106 mol PbS × (3 mol O2 / 2 mol PbS) = 0.0159 mol O2.

Since the actual amount of O2 we have is less than what is needed (0.0159 mol), O2 is the limiting reactant.

This means that PbS is in excess and we can calculate the mass of lead produced using the amount of O2 that reacted.

The balanced equation tells us that 3 moles of O2 produce 2 moles of lead.

Therefore,0.0588 mol O2 × (2 mol Pb / 3 mol O2) = 0.0392 mol PbFinally, we can calculate the mass of lead produced using the number of moles and the molar mass of lead.mass of Pb = number of moles × molar mass= 0.0392 mol × 207.2 g/mol= 8.12 g.

Therefore, 8.12 g of lead will be produced if 2.54 g of PbS is burned with 1.88 g of O2.

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The following are categorized into polar or nonpolar molecules:

a) Beryllium chloride (BeCl₂) is a nonpolar molecule. The Be-Cl bond is polar due to the electronegativity difference between beryllium and chlorine, but the molecule is linear with the two polar bonds pointing in opposite directions, resulting in a net dipole moment of zero.

b) Hydrogen sulphide (H₂S) is a polar molecule. The H-S bond is polar due to the electronegativity difference between hydrogen and sulfur, and the molecule has a bent shape, resulting in a net dipole moment that is not zero.

c) Sulphur trioxide (SO₃) is a nonpolar molecule. The S-O bonds are polar due to the electronegativity difference between sulfur and oxygen, but the molecule is trigonal planar with the three polar bonds pointing in different directions, resulting in a net dipole moment of zero.

d) Water (H₂O) is a polar molecule. The H-O bond is polar due to the electronegativity difference between hydrogen and oxygen, and the molecule has a bent shape, resulting in a net dipole moment that is not zero.

e) Trichloromethane (CHCl₃) is a polar molecule. The C-Cl bonds are polar due to the electronegativity difference between carbon and chlorine, and the molecule has a tetrahedral shape, resulting in a net dipole moment that is not zero.

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Answer:

178.56 cm3

Explanation:

V1/T1=V2/T2

V2=V1T1/T2

V1 (volume) is 200 cm3

T1 (temperature) is 60°C but convert into Kelvin

T2 (temperature) is 100 °C but convert into Kelvin

Plug in the values...

V2= (200 cm3)(60+273.15)/ (100+273.15)

V2=178.56 cm3

Answer:

178.56 cm3

Explanation:

Answer:

0.521 moles still present in the container.

Explanation:

It is possible to answer this question by using the general gas law, that is:

PV = nRT

Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)

Replacing with values of the initial conditions of the container, its volume is:

V = nRT / P

V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm

V = 12.9L

When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and the volume of the container still constant. Again, using general gas law:

PV / RT = n

0.998atm*12.9L / 0.082atmL/molK*301.25K = n

0.521 moles = n

Ideal gas law is the hypothetical equation in which the pressure, volume, and temperature of the gas are directly related. It can be denoted as:

PV = nRT

The number of moles still present in the container is 0.521.

The ideal gas law is:

PV = nRT

where,

P = Pressure

V = Volume

R = Gas constant

T =Temperature

n = moles

Given:

Moles in container = 2.00

Temperture = 294.85 K

Pressure = 3.75 atm

Volume =?

Substituting the values:

V = nRT/P

\(\text V&=\dfrac{2\times0.082 \times294}{3.75}\)

Volume = 12.9 L

Now, the condition when changed, such that temperature is 301.25 K, pressure is 0.998 atm, and Volume is 12.9 L, then moles will be equal to:

\(\begin{aligned}\dfrac{\text{PV}}{\text{RT}}&=\text n\\\\\dfrac{0.998 \times 12.9} {0.082 \times 301.25}&=\text n\end\)

n = 0.521 moles

Therefore, 0.521 moles is still present in the container.

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Answer:

Can you translate it on English so we can understand

Answer:

a process that involves rearrangement of the molecular or ionic structure of a substance, as distinct from a change in physical form or a nuclear reaction

Answer:

mole fraction of C6H6 = 0.613 atm

Explanation:

The equation for this reaction is :

          \(C_6H_6 _{(g)} + 3H_2_{(g)} \to C_6H_{12}_{(g)}\)

Initial      P₁            P₂             0

Final        0           P₂ -P₁/2      P₁

After completion of the reaction;

P₁  +  P₂  = 1.21 atm                    ----- (1)

P₂  -  P₁/2 +  P₁ = 0.839 atm

P₂ + P₁/2 = 0.839 atm          ----- (2)

Subtracting (2) from (1); we have:

P₁/2 = 0.371

P₁ = 0.742 atm

From(1)

P₁  +  P₂  = 1.21 atm

0.742 atm + P₂  = 1.21 atm

P₂  = 1.21 atm - 0.742 atm

P₂  = 0.468 atm

Thus, the partial pressure of C6H6 = 0.742 atm

∴

Partial pressure = Total pressure × mole fraction of C6H6

mole fraction of C6H6 = Partial pressure /  Total pressure

mole fraction of C6H6 = 0.742 atm / 1.21 atm

mole fraction of C6H6 = 0.613 atm

I apologize of I'm wrong but its probably OA

Sorry once again!

~wise rat

Answer: C.

Explanation:

gasoline evaporating into the air

Answer: The enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

The final enthalpy will be:

The enthalpy of the second intermediate equation is halved and has its sign changed.

First reaction:

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction:

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Overall reaction:

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is the sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

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Ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)) mixture could be a useful buffer in a solution. Option C

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer system works by the principle of Le Chatelier's principle, where the equilibrium is shifted to counteract the changes caused by the addition of an acid or a base.

In option A, acetic acid (\(CH_3CO_2H\)) is a weak acid, but hydrochloric acid (HCl) is a strong acid. This combination does not form a buffer because HCl is completely dissociated in water and cannot provide a significant concentration of its conjugate base.

Option B consists of sodium hydroxide (NaOH), which is a strong base, and elemental sodium (Na), which is a metal. This combination does not form a buffer as there is no weak acid-base pair involved.

Option D contains acetic acid (\(CH_3CO_2H\)), a weak acid, and ammonia (\(NH_3\)), a weak base. Although they are weak acid and base, they do not form a buffer system together as they are both weak acids or bases and lack the required conjugate acid-base pair.

Option C, ammonia (\(NH_3\)), is a weak base, and ammonium chloride (\(NH_4Cl\)) is its conjugate acid. This combination can form a buffer system. When ammonia reacts with water, it forms ammonium ions (NH4+) and hydroxide ions (OH-).

The ammonium ions act as the weak acid, while the ammonia acts as the weak base. The addition of a small amount of acid will be counteracted by the ammonium ions, and the addition of a small amount of base will be counteracted by the ammonia, thus maintaining the pH of the solution relatively stable.

Therefore, option C, consisting of ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)), is the suitable mixture that could be a useful buffer in a solution.

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The molecular or formula mass of each of the given substances is given below:

(a) P₄ = 128 amu

(b) H₂O = 18 amu

(c) Ca(NO₃)₂ = 124 amu

(d) CH₃CO₂H (acetic acid) = 60 amu

(e) C₁₂H₂₂O₁₁ (sucrose, cane sugar) = 342 amu

The molecular or formula mass of a substance id the sum of the masses of the atoms of the elements present in that substance.

The molecular or formula mass of each of the given substances is determined as follows:

(a) P₄ =  32 * 4

P₄ = 128 amu

(b) H₂O = 2 * 1 + 16

H₂O = 18 amu

(c) Ca(NO₃)₂ = 40 + (14 * 2 + 16 * 3 * 2)

Ca(NO₃)₂ = 124 amu

(d)  CH₃CO₂H = 2 * 12 + 4 * 1 + 16 * 2

CH₃CO₂H = 60 amu

(e) C₁₂H₂₂O₁₁  = 12 * 12 + 1 * 22 + 16 * 11

C₁₂H₂₂O₁₁ = 342 amu

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Answer:

7.8%

Explanation:

Given that:

The initial mass amount of aspirin = 0.020g

The standard molar mass of aspirin = 240 g/mol

Thus, the number of moles = mass/molar mass

= 0.020/240

= 0.0000833 moles

Now, the molarity of aspirin in the solution(diluted) \(C_1\) = \(\dfrac{8.33 \times 10^{-5}}{0.01}\)

= \(= 8.33 \times 10^{-3} \ M\) (provided the volume v = 0.01 L)

The absorbance of the sample solution A =1.07

The path length (b) = 1 cm

From the standard value of salicylic acid, the coefficient (e)= 139.322 /M/cm

Now; according to Beer's law, the concentration of aspirin is:

A = e×b×c

\(c = \dfrac{A}{eb}\)

\(c = \dfrac{1.07}{139.322 \times 1}\)

c = 0.00768 M

Finally, relating the concentration of the aspirin, the percentage of salicylic acid the product  \(= \dfrac{0.00833-0.00768}{0.00833} \times 100\)

= 7.8%

The amount that could cause death in a 150 pound person is 675000 mg.

We define the LD 50 as the amount of the substance that could be used to cause mortality in about 50 percent of the population that is undergoing the test. In this case, we have been told that the  LD50 for Allura Red ingested for rats and rabbits is 10,000 mg/kg  an now we are asked for the LD 50 in a person that has a weight of 150-lb.

The first step in the work is to convert the mass of the person from pound to kilogram from here;

Given that

1 pound =  0.45 kilo gram

150 pounds = 150 pounds * 0.45 kilo gram/1 pound

= 67.5 Kg

Now, If

10,000 mg causes mortality in 1 Kg

x mg would cause mortality in 67.5 Kg

x = 675000 mg

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The net ionic equation is applied mainly in neutralization reactions, double displacement reactions and redox reactions. The net ionic equation is as follows

Li⁺(aq)+Cl⁻(aq)\(\rightarrow\)LiCl(s)

The net ionic equation represents a chemical equation of a reaction that give an idea of species that are really participating in the reaction. We can write net ionic equations for those electrolyte which are strong in water.

The balanced chemical equation for the given reaction between lithium fluoride and potassium chloride

LiF(aq) + KCl(aq) \(\rightarrow\) KF(aq) +LiCl

The ionic equation

Li⁺ + F⁻+K⁺+Cl⁻\(\rightarrow\)LiCl+ K⁺+F⁻

Cancelling the spectator ions both sides

The net ionic equation is

Li⁺(aq)+Cl⁻(aq)\(\rightarrow\)LiCl(s)

Therefore the net ionic equation for the given chemical reaction is Li⁺(aq)+Cl⁻(aq)\(\rightarrow\)LiCl(s)

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Answer:

Small size

Explanation:

Answer:

C. The wind picks up speed

Explanation:

Im not sure but i hope im correct

Btw goodluck:)

Answer:

waves, energy is transferred through vibrations of electric and magnetic fields.

Explanation:

In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles

By removing the need to build additional transmission lines and equipment, energy storage may reduce costs for utilities and their customers.

By removing the need to build additional transmission lines and equipment, energy storage may reduce costs for utilities and their customers. Energy storage's inherent ability to offer backup power in the event of grid failure is a feature that both residential consumers and commercial owners find highly desirable.

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Answer:

2044

Explanation:

Explanation:

A nitrogen-containing compound that shows no absorption band at around 3400 cm^−1 and no absorption bands between approximately 1700 cm^−1 and 1600 cm^−1 is likely an amide compound.

Amides typically exhibit a characteristic absorption band in the region of 3200-3500 cm^−1 due to the N-H stretching vibration. The absence of this absorption band suggests the absence of N-H bonds, which rules out compounds like primary or secondary amines.

The absence of absorption bands between 1700 cm^−1 and 1600 cm^−1 eliminates functional groups such as carbonyl compounds (e.g., aldehydes, ketones, carboxylic acids, esters) and imines, which typically exhibit absorption in this region.

Therefore, based on the given information, it can be inferred that the compound is likely not an amine, carbonyl compound, or imine. Other classes of compounds that do not possess these characteristic absorption bands would need to be considered.

Magnesium bromide is the limiting reagent, and the amount of \(MgCl_{2}\) formed will be limited by the amount of magnesium bromide available. The excess reactant will be sodium chloride.

What is limiting reagent?

To determine which of the two reactants, magnesium bromide or sodium chloride, is the limiting reagent, we need to calculate the amount of product that can be formed from each reactant and compare the results.

Let's assume that magnesium bromide and sodium chloride react to form magnesium chloride and sodium bromide, and the balanced chemical equation for the reaction is:

\(MgBr_{2}\) + 2NaCl → 2NaBr + \(MgCl_{2}\)

The molar masses of each compound are:

The first step is to calculate the number of moles of each reactant, given that we have 2 grams of each:

moles of \(MgBr_{2}\) = 2 g / 184.11 g/mol = 0.01085 mol

moles of NaCl = 2 g / 58.44 g/mol = 0.03423 mol

Now, we can use the stoichiometry of the balanced equation to calculate the theoretical yield of product for each reactant:

Using \(MgBr_{2}\): 0.01085 mol \(MgBr_{2}\) x (1 mol \(MgCl_{2}\) / 1 mol \(MgBr_{2}\)) x (95.21 g \(MgCl_{2}\) / 1 mol \(MgCl_{2}\)) = 1.034 g \(MgCl_{2}\)

Using NaCl: 0.03423 mol NaCl x (1 mol \(MgCl_{2}\) / 2 mol NaCl) x (95.21 g \(MgCl_{2}\) / 1 mol \(MgCl_{2}\)) = 1.633 g \(MgCl_{2}\)

From the above calculation, we can see that the theoretical yield of \(MgCl_{2}\) is lower when magnesium bromide is used as the reactant (1.034 g), compared to when sodium chloride is used (1.633 g). Therefore, magnesium bromide is the limiting reagent, and the amount of \(MgCl_{2}\) formed will be limited by the amount of magnesium bromide available. The excess reactant will be sodium chloride.

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Complete question is: if given 2 grams of magnesium bromide and sodium chloride; magnesium bromide would be the limiting reagent.

The limiting reagent would be the magnesium bromide because it is present in the smallest quantity.

Magnesium is a silvery-white chemical element and is the ninth most abundant element in the universe. It is a key component of chlorophyll and is essential for plant and animal life. Magnesium is also an important mineral for human health, playing a role in over 300 biochemical processes in the body. It is involved in energy production, DNA and RNA synthesis, and muscle, nerve, and enzyme function. Magnesium is found in many foods, including green leafy vegetables, nuts, seeds, and whole grains. Magnesium supplements are also available and are commonly used to help prevent or treat magnesium deficiency.

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Answer:

c

Explanation:

electrons

The electrons of an atom involved in the formation of a chemical bond are called valence electrons. Therefore, option (A) is correct.

Valence electrons can be defined as the electrons present in the outermost shell of an atom while the electrons occupied in the inner shell are called core electrons. Lewis structures can be helpful to determine the number of valence electrons and know the types of chemical bonds.

Valence electrons are distributed in different shells and these electrons are cause interaction between different atoms and responsible for the formation of chemical bonds between atoms to complete their octet.

Only electrons filled in the outermost shell can able participate in the formation of a chemical bond or a molecule and decide the reactivity of the element.

For example, the elements of group 1 have one electron in their valence shell. They have a great tendency to lose that electron to acquire the configuration of the nearest noble gas.

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The answer is 19.2 grams.

Since it asks for three significant figures, all you need to do is count three digits and that should give you your final answer after rounding.

Answer:

2.9 psi

Explanation:

Pressure (mmHg) = 150 mmHg

Pressure (psi) =?

The pressure in pound per square inch (psi) can be obtained as follow:

51.715 mmHg = 1 psi

Therefore,

150 mmHg = 150 mmHg × 1 psi / 51.715 mmHg

150 mmHg = 2.9 psi

Thus, pressure in pound per square inch (psi) is 2.9 psi.

Number of molecules in 1 dm³ Oxygen = 2.71 x 10²²

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

The mole is the number of particles contained in a substance

1 mol = 6.02.10²³

1 dm³ of oxygen = 1 L Oxygen

\(\tt \dfrac{1}{22.4}=0.045`mol\)

n=mol=0.045

No = 6.02.10²³

\(\tt N=n\times No\\\\N=0.045\times 6.02\times 10^{23}\\\\N=2.71\times 10^{22}\)

Answer: H2O + CO2 --> H2CO3

Explanation:

Water and Carbon Dioxide react to form Carbonic Acid

H2O + CO2 --> H2CO3

Standard state of the given equation will be:

Pb (s) + N2(g) + 3O2(g) --> Pb(NO3)2(s)

A chemical equation is a word used frequently in everyday chemistry. Jean Beguin, a French chemist, developed the first chemical equations in 1615. A chemical equation is simply a written representation of the process that occurs during a chemical reaction using numbers and symbols.

Explain:

Pb (NO3)2(s)

elements:

Pb, N, O

elemental states

Pb (s)   N2(g) , O2(g)

equation

Pb (s) + N2(g) + 3O2(g) --> Pb(NO3)2(s)

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