What is the hydrogen-ion concentration of the ph is 3.7

Answer:

Check the explanation

Explanation:

Lorenz force exerted on electron by magnetic field is

F=e*v*H, where e=1.602E-19 C, H=3.64E-3 T, v is speed of electron;

? meanwhile Lorenz force is centripetal force F=m*v^2/r, where mass of electron m=9.11E-31 kg, r is radius of the path of electron;

? therefore F=F; e*v*H = m*v^2/r; eH=m*(v/r), hence

v/r = eH/m =w is angular speed of electron, hence

T=2pi/w =2pi*m/(eH) is period of rotation of electron;

e- and e+ starting at the same point are moving in the same circular path and in opposite directions, and should meet in T/2 time;

T/2 = pi*m/(eH) = pi*9.11E-31 /(1.602E-19 *3.48E-3) =5.13E-9 s;

Answer:

D)

Explanation:

I just answered the question and got it right.

Answer:

He added 992 milimoles of zinc nitrate (Zn(NO3)2)

Explanation:

Step 1: Data given

Volume of zinc nitrate = 435.0 mL =0.435 L

Molarity of zinc nitrate = 2.28 M

Step 2: Calculate moles zinc nitrate

Moles = molarity*  volume

Moles Zn(NO3)2 = 2.28 M * 0.435 L

Moles Zn(NO3)2 = 0.9918 moles

Step 3: Convert moles to milimoles

Moles Zn(NO3)2 = 0.9918 moles

Moles Zn(NO3)2 = 0.9918 * 10^3 milimoles

Moles Zn(NO3)2 = 991.8 milimoles ≈ 992 milimoles

He added 992 milimoles of zinc nitrate (Zn(NO3)2)

Final answer:

To determine the millimoles of zinc nitrate added, convert 435.0 mL to liters, use the molarity of 2.28 M to find moles, and then multiply by 1000 to get 992.4 millimoles.

Explanation:

To calculate the millimoles of zinc nitrate a chemist has added to the reaction flask, you first need to know the concentration of the solution and the volume of the solution used. Here, we have a 2.28 M zinc nitrate solution, and the volume used is 435.0 mL. To find the millimoles, first convert the volume from milliliters to liters (since molarity is moles per liter), then use the molarity to find the moles of zinc nitrate, and finally convert that to millimoles.

Convert volume to liters: 435.0 mL × (1 L / 1000 mL) = 0.435 L

Calculate moles of Zn(NO3)2: 2.28 moles/L × 0.435 L = 0.9924 moles

Convert moles to millimoles: 0.9924 moles × 1000 mmol/mole = 992.4 mmol

Therefore, the chemist has added 992.4 millimoles of zinc nitrate to the flask.

Answer:

see below

Explanation:

for A + 2B => Products ...

Rate Law => Rate =k[A][B]ˣ

As shown in expression, A & B are included, C is not.

Answer:

Exponent of a is 1, exponent of b is 2 and exponent of c = 0

Explanation:

or the rate equation is the expression which relates the rate of the reaction with the concentration of pressure of the reactants. The rate law is expressed in terms of the molar concentration of the reactants with each term raised to power its stoichiometric coefficient.

The given question is incomplete. The complete question is:

Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 1.60 g of sodium chloride is produced from the reaction of 1.8 g of hydrochloric acid and 1.4 g of sodium hydroxide, calculate the percent yield of sodium chloride. Be sure your answer has the correct number of significant digits in it.

Answer: Thus the percent yield of sodium chloride is 78.0%

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} HCl=\frac{1.8g}{36.5g/mol}=0.049moles[/tex]

[tex]\text{Moles of} NaOH=\frac{1.4g}{40g/mol}=0.035moles[/tex]

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]

According to stoichiometry :

1 mole of [tex]NaOH[/tex] require = 1 mole of [tex]HCl[/tex]

Thus 0.035 moles of [tex]NaOH[/tex] will require=[tex]\frac{1}{1}\times 0.035=0.035moles[/tex] of [tex]HCl[/tex]

Thus [tex]NaOH[/tex] is the limiting reagent as it limits the formation of product and [tex]HCl[/tex] is the excess reagent.

As 1 mole of [tex]NaOH[/tex] give = 1 mole of [tex]NaCl[/tex]

Thus 0.035 moles of [tex]NaOH[/tex] give =[tex]\frac{1}{1}\times 0.035=0.035moles[/tex]  of [tex]NaCl[/tex]

Mass of [tex]NaCl=moles\times {\text {Molar mass}}=0.035moles\times 58.5g/mol=2.05g[/tex]

[tex]{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%[/tex]

[tex]{\text {percentage yield}}=\frac{1.60g}{2.05g}\times 100\%=78.0\%[/tex]

Thus the percent yield of sodium chloride is 78.0%

Final answer:

To find the empirical formula of the gaseous hydrocarbon, assume 100 grams of the compound. Since it is 11% hydrogen by mass, it contains 11 grams of hydrogen. The remaining mass is carbon. Divide the moles of each element by the smallest number of moles to get the empirical formula CH₂. The molecular formula cannot be determined without the molar mass.

Explanation:

Empirical Formula Calculation

To find the empirical formula, assume 100 grams of the gaseous hydrocarbon. Since it is 11% hydrogen by mass, this means it contains 11 grams of hydrogen. The remaining mass, 89 grams, must be carbon. The molar mass of hydrogen is 1 g/mol, and the molar mass of carbon is 12 g/mol. Divide the moles of each element by the smallest number of moles to get the empirical formula. In this case, 11 g H / 1 g/mol = 11 mol H and 89 g C / 12 g/mol = 7.42 mol C. Dividing both by the smallest number of moles, we get the empirical formula CH₂.

To determine the molecular formula, you'll need the molar mass of the compound. However, this information is not provided in the question. Without the molar mass, it is not possible to determine the molecular formula.

To find the final molarity of bromide ions in a solution, we need the mass of the dissolved potassium bromide and the volume of the solution. Then we do a simple molar calculation. The molarity of KBr and Br- are equivalent as KBr disassociates fully in solution.

In order to determine the final molarity of the bromide anion, we first need to know the amount of potassium bromide (KBr) dissolved and the volume of the solution it's dissolved in. However, these key data are missing in the question.

Regardless, the steps to calculate molarity (M) are as follows:

In this particular scenario, we also know that one mole of KBr results in one mole of bromide ions (Br-), as it disassociates fully upon dissolution. So, the molarity of KBr would be equal to the molarity of Br-.

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Answer:

d. I and II

Explanation:

In chemistry, it is an old analytical technique to determine the amount of free chloride ions by silver chloride precipitation. This precipitation method could both be used to determine the initial chloride ion concentration in different analyte solutions or to determine the number of uncoordinated chloride ions in a complex. The number of moles of chloride ions in the silver chloride, shows the number of uncoordinated ions in a chloride ion containing complex.

The concentration of chloride ions in an analyte could be gravimetrically determined by silver chloride precipitation.

Answer:

P = 2.25 atm

Explanation:

Given data:

Mass of dry ice = 10 g

Volume of container = 2.5 L

Temperature = 25°C (25+273=298 K)

Pressure inside container = ?

Solution:

First of all we will calculate the number of moles of dry ice.

Number of moles = mass / molar mass

Number of moles = 10 g/ 44 g/mol

Number of moles = 0.23 mol

Now we will determine the pressure.

PV = nRT

P = nRT/V

P = 0.23 mol × 0.0821 atm.L/mol.K  × 298 K / 2.5 L

P = 5.63 atm  / 2.5

P = 2.25 atm

Answer:

The answer is 4.8659

Explanation:

Given:

[C₆H₅COOH]=0.11M

[C₆H₅COO]=2*0.2=0.4M

Ka=6.3x10⁻⁵

First, calculate the pKa:

[tex]pKa=-logKa=-log(6.3x10^{-5} )=4.2007[/tex]

The pH is:

[tex]pH=pKa+log\frac{C6H5COO]}{[C6H5COOH]} =4.2007+log\frac{0.4}{0.11} =4.7614[/tex]

Like the volume is 5L, the volume of C₆H₅COO is x, then, the volume of C₆H₅COOH is 5-x

[tex]4.7614=4.2007+log\frac{0.4x}{0.11*(5-x)}[/tex]

[tex]0.5607=log\frac{0.4x}{0.11*(5-x)}[/tex]

Solving for x:

x=2.49L=2490mL of C₆H₅COO

2510mL of C₆H₅COOH

The milimoles of C₆H₅COOH and C₆H₅COO is:

nC₆H₅COOH=(0.11*2510)-50=226.1mmol

nC₆H₅COO=(0.4*2490)+50=1046mmol

The pH is:

[tex]pH=4.2007+log\frac{1046}{226.1} =4.8659[/tex]

Answer:

4.86

Explanation:

The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.

Please kindly check attachment for the step by step solution of the given problem.

The molar mass or gram formula mass is the conversion factor used to convert between grams and moles in chemistry.

The conversion factor used to convert between grams and moles in chemistry is called the molar mass or the gram formula mass. It is defined as the mass of one mole of a substance. To convert grams to moles, you divide the given mass by the molar mass of the substance. To convert moles to grams, you multiply the given moles by the molar mass.

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Answer:

A mixture is made when two or more substances are combined, but they are not combined chemically. A chemical reaction is a transformation from one set of chemicals into another set.

Explanation:

A mixture can be homogeneous or heterogeneous, when homo, you cannot see each variable but hey are there. Hetero, means you can see each variable in the mixture. It is considered a mixture and not a chemical reaction because it can be reversed. It will always be able to go back to its separate forms and keep its original composition. When there is a chemical reaction, it is hard to separate and go back to the original variables because there is a molecular bond.

Answer: starting on the left:

1. 0.1M HCI

2. 0.001M HCI

3. 0.00001M HCI

4. distilled water

5. 0.00001 NaOH

6. 0.001M NaOH

7. 0.1M NaOH

Explanation:

The solutions arranged from most acidic to most basic would be: 0.1 M HCl, 0.001 M HCl, 0.00001 M HCl, Distilled Water, 0.00001 M NaOH, 0.001 M NaOH, and finally 0.1 M NaOH.

The pH of a solution depends on the concentration of hydronium ions (H+) and hydroxide ions (OH-). In this case, a lower concentration means less acidity or basicity. The solutions arranged in order from most acidic to most basic would be:

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The Ksp of the solution is [tex]1.607*10^-^1^4[/tex]

Data;

This is the point or temperature in which a solute is completely soluble in a solvent.

The solubility of iron(ii) hydroxide can be calculated by using the equation of reaction.

[tex]Fe(OH)_2 \to Fe^2^+ + 2OH^-\\[/tex]

But the solubility of the solution is given as 1.59*10^-5 mol/L

[tex][Fe^2^+] = s\\\\[/tex]

[tex][OH^-] = 2s\\K_s_p = s * (2s)^2 = 4s^3[/tex]

Let's substitute the values and solve

[tex]K_s_p = 4 *( 1.59*10^-^5)^3 = 1.607*10^-^1^4[/tex]

The Ksp of the solution is [tex]1.607*10^-^1^4[/tex]

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Final answer:

The Ksp of iron(II) hydroxide at 25 °C is calculated to be 1.61 x 10−14, based on its given solubility and the stoichiometry of its dissociation in water.

Explanation:

The solubility product constant, Ksp, for iron(II) hydroxide, Fe(OH)2, at 25 °C can be calculated using the mol/L solubility given and the stoichiometry of the dissolution reaction:

Fe(OH)2 (s) → Fe2+ (aq) + 2OH− (aq)

For iron(II) hydroxide, the solubility is 1.59 x 10−5 mol/L. This is the concentration of Fe2+. Since the ratio of Fe2+ to OH− is 1:2, this means the concentration of OH− is twice this value, thus 3.18 x 10−5 mol/L. The Ksp is the product of these ion concentrations:

Ksp = [Fe2+][OH−]2 = (1.59 x 10−5)(3.18 x 10−5)2

Calculating the above expression:

Ksp = 1.59 x 10−5 x (1.01 x 10−9)

Ksp = 1.61 x 10−14

This is the Ksp of iron(II) hydroxide at 25 °C, expressed in scientific notation to two significant figures.

This is an incomplete question, here is a complete question.

The concentration of Cu²⁺ ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na₂S)solution to 0.700 L of the water.

The molecular equation is:

[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]

Write the net ionic equation and calculate the molar concentration of Cu²⁺ in the water sample if 0.0177 g of solid CuS is formed.

Answer :

The net ionic equation will be,

[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]

The concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The molecular equation is:

[tex]Na_2S(aq)+CuSO4(aq)\rightarrow Na_2SO_4(aq)+CuS(s)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]2Na^+(aq)+S^{2-}(aq)+Cu^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CuS(s)+2Na^+(aq)+SO_4^{2-}(aq)[/tex]

In this equation, [tex]Na^+\text{ and }SO_4^{2-}[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Cu^{2+}(aq)+S^{2-}(aq)\rightarrow CuS(s)[/tex]

Now we have to calculate the mass of [tex]CuSO_4[/tex]

Molar mass of [tex]CuSO_4[/tex] = 159.5 g/mol

Molar mass of CuS is = 95.5 g/mol

From the balanced chemical reaction we conclude that,

As, 95.5 g of CuS  produces from 159.5 g [tex]CuSO_4[/tex]

As, 0.0177 g of CuS  produces from [tex]\frac{159.5}{95.5}\times 0.0177=0.0296g[/tex] [tex]CuSO_4[/tex]

Now we have to calculate the concentration of [tex]CuSO_4[/tex]

[tex]\text{Concentration}=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4\times \text{Volume of solution (in L)}}[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Concentration}=\frac{0.0296g}{159.5g/mol\times 0.700L}=2.65\times 10^{-4}M[/tex]

Concentration of [tex]Cu^{2+}[/tex] = [tex]2.65\times 10^{-4}M[/tex]

Therefore, the concentration of [tex]Cu^{2+}[/tex] is, [tex]2.65\times 10^{-4}M[/tex]

Answer:

[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.

Explanation:

K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)

PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)              

The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.

Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which  means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.

Assuming all PbCrO₄ formed in the 175 ml solution remained ionized,  then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.

To test if saturation occurs, Qsp must be > than Ksp ...

=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 => This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation. That is ...

          PbCrO₄(s) ⇄  Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³

C(i)           ---               0.00M      0.0336M

ΔC           ---                  +x                +x          

C(eq)       ---                    x           0.0336 + x ≅ 0.0336M

Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³

∴[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.

Answer:

[Pb⁺²] remaining = 9.0 x 10⁻¹²M in Pb⁺² ions.

Explanation:

K₂CrO₄(aq) + Pb(NO₃)₂(aq) => 2KNO₃(aq) + PbCrO₄(s)

PbCrO₄(s) ⇄ Pb⁺²(aq) + CrO₄⁻²(aq)              

The K₂CrO₄(aq) + Pb(NO₃)₂(aq) is 1:1 => moles K₂CrO₄ added = moles of Pb(NO₃)₂ converted to PbCrO₄(s) in 175 ml aqueous solution.

Given rxn proceeds until [CrO₄⁻] = 0.0336 mol/L which  means that moles Pb(NO₃)₂ converted into PbCrO₄ is also 0.0336 mol/L.

By assumption if we say, all PbCrO₄ formed in the 175 ml solution remained ionized,  

Then [Pb⁺²] = [CrO₄⁻²] = 0.0336mol/0.175L = 0.192M in each ion.

To test if saturation occurs,

Qsp must be > than Ksp ...

=> Qsp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = (0.0.192)² = 0.037 >> Ksp(PbCrO₄) = 3 x 10⁻13 =>

This means that NOT all of the PbCrO₄ formed will remain in solution. That is, the ppt'd PbCrO₄ will deliver Pb⁺² ions into solution until saturation is reached. However, one should note that determination of the concentration of Pb⁺² delivered back into solution will be in the presence of 0.0366M CrO⁻² ions and the problem then becomes a common ion type calculation.

i.e .........

PbCrO₄(s) ⇄  Pb⁺²(aq) + CrO₄⁻²(aq); Ksp = 3 x 10⁻¹³

C(i)           ---       0.00M      0.0336M

ΔC           ---       +x                +x          

C(eq)       ---        x          

0.0336 + x ≅ 0.0336M

Ksp = [Pb⁺²(aq)][CrO₄⁻²(aq)] = [Pb⁺²(aq)](0.0336M) = 3 x 10⁻¹³

Therefore;

[Pb⁺²(aq)] = (3 x 10⁻¹³/0.0336)M = 8.93 x 10⁻¹²M ≅ 9.0 x 10⁻¹²M in Pb⁺² ions.

Question: The question is not complete. Find below the complete question and the answer.

Alab group was supposed to make 14 mL of a 36% acid solution by mixing a 20% solution, a 26% solution, and a 42% solution. However, the 20% solution was mislabeled, and was actually a 10% solution, so the lab group ended up with 14 mL of a 34% acid solution, instead. If the augmented matrix that represents the system of equations is given below, what are the volumes of the solutions that should have been mixed? mL

Volume of 20% solution= ?

Volume of 26% solution = ?

Volume of 42% solution= ?   Round to the nearest whole number ml

Answer:

Volume of 20% solution= 3 mL

Volume of 26% solution = 1 mL

Volume of 42% solution= 10 mL

Explanation:

Find attached of the calculations.

Answer:

a. 0.8125 moles of sucrose

b. 277.8 g of sucrose

Explanation:

Consider these relation's value:

Molarity = Mol / Volume(L)

Mol = Molarity . Volume(L)

Volume(L) = Mol / Molarity

So, Molarity = 3.25M

Volume(L) = 0.25L

Molarity . Volume(L) = Mol → 3.25mol/L . 0.25L = 0.8125 mol

Let's convert the moles to mass → 0.8125 mol . 342 g /1mol = 277.8 g of sucrose

Answer:

We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose

Explanation:

Step 1: Data given

Molarity of a sucose solution = 3.25 M

Molar mass of sucrose = 342 g/mol

Step 2:  Calculate moles sucrose

Moles sucrose = molarity sucrose solution * volume solution

Moles sucrose = 3.25 M * 0.25 L

Moles sucrose =  0.8125 moles sucrose

Step 3: Calculate mass of sucrose

Mass sucrose = moles sucrose * molar mass sucrose

Mass sucrose = 0.8125 moles * 342 g/mol

Mass sucrose = 277.9 grams

We have 0.8125 moles sucrose in this solution. This is 277.9 grams of sucrose

Answer:

the concentration of sulfuric acid in the flask is  0.375 M

Explanation:

H2SO4 (aq) + 2 NaOH (aq) → 2 H2O (l) + Na2SO4 (aq)  

Moles of NaOH = 15 x 0.25 /1000

                        = 0.00375

Moles of H2SO4 needed to neutralize = 0.00375 /2

                       = 0.001875

Molarity of H2SO4

                      = 0.001875 x 1000 /20

                      = 0.09375 M

concentration of sulfuric acid in the flask 0.09375 M

Moles

V2 = 20 ml

N2 =?

Substitute in the equation and find N2

N2 = 2 x 15 x 0.25 / 20

     = 0.375 M

Thus, the concentration of sulfuric acid in the flask is  0.375 M

Answer:

Answer: (1R,2S) / (1S, 2R) , (1R,2R) / (1S, 2S)  

Explanation:

Sodium borohydride reduction of benzoin will give four possible stereo isomers out of which are (1R,2S) - (1S, 2R) isomers and (1R,2R) - (1S, 2S) isomers which are known as enantiomers.

In general enantiomers show single spot in the TLC as they do not show any difference in Rf value (i.e) (1R,2S) - (1S, 2R) isomers show only one spot although they are two compounds and also (1R,2R) - (1S, 2S) isomers also show one spot. That is the reason why you are observing two spots in the TLC ( of reaction mixture) other than starting materilal.

Answer:

Explanation:

find the solution below

Answer:

the pressure of hydrogen gas in the flask is 725 torr

Explanation:

the solution is in the attached Word file

The pressure of hydrogen gas in the flask is [tex]\( 725.6 \, \text{torr} \)[/tex].

To determine the pressure of hydrogen gas collected over water, we need to account for the vapor pressure of water at the given temperature. The total pressure in the flask is the sum of the partial pressures of hydrogen gas and water vapor.

The steps to solve this are:

1. Find the vapor pressure of water at the given temperature (23.5°C):

The vapor pressure of water at 23.5°C is approximately 21.1 torr (this value can be found in tables of water vapor pressures).

2. Calculate the pressure of hydrogen gas:

The total pressure in the flask is the atmospheric pressure, which is 746.7 torr. The pressure of hydrogen gas is the total pressure minus the vapor pressure of water.

[tex]\[\text{Pressure of hydrogen gas} = \text{Total pressure} - \text{Vapor pressure of water}\][/tex]

[tex]\[P_{\text{H}_2} = 746.7 \, \text{torr} - 21.1 \, \text{torr}\][/tex]

[tex]\[P_{\text{H}_2} = 725.6 \, \text{torr}\][/tex]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

          A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600g of water. From the given information, the final temperature of the reaction is 22.71° C

From the given parameters:

From the listed parameters, the first step is to determine the number of moles of the sample by using the relation:

[tex]\mathbf{number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

[tex]\mathbf{number \ of \ moles = \dfrac{0.500 \ g}{213 \ g/mol}}[/tex]

number of moles of C7H5N2O6 = 0.00235 moles

However, the heat of combustion is the amount or quantity of heat energy released when one mole of a sample is burned.

Mathematically;

[tex]\mathbf{\Delta H_c =\dfrac{q_{rxn} }{ n}}[/tex]

where;

[tex]\mathbf{ q_{rxn}}[/tex] = quantity of heat released

n = number of moles

Making  [tex]\mathbf{ q_{rxn}}[/tex] the subject of the formula:

[tex]\mathbf{ q_{rxn}= \Delta H_{rxn} \times n }[/tex]

[tex]\mathbf{ q_{rxn}=-3374 \ kJ/moles \times 0.00235 \ moles }[/tex]

[tex]\mathbf{ q_{rxn}=-7.9289 \ kJ }[/tex]

[tex]\mathbf{ q_{rxn}=-7928.9 \ J }[/tex]

The heat released [tex]\mathbf{ q_{rxn}}[/tex] = heat absorbed by bomb calorimeter + heat absorbed by water

∴

[tex]\mathbf{ q_{rxn}= -(m\times c_{bc} \Delta T + c_{water} \times \Delta T)}[/tex]

[tex]\mathbf{-7928.9 \ J = -( 420 J/^0C \times \Delta T + 600 \ g \times 4.18 \ J/g ^0 C\times \Delta T)}[/tex]

[tex]\mathbf{7928.9 = (2928 \Delta T )}[/tex]

[tex]\mathbf{ \Delta T = \dfrac{7928.9}{2928} }[/tex]

[tex]\mathbf{ \Delta T =2 .71 ^0 \ C}[/tex]

Since ΔT = [tex]\mathbf{T_2 - T_1}[/tex]

2.71° C = T₂ - 20° C

T₂ = 20° C + 2.71° C

T₂ = 22.71° C

Therefore, we can conclude that the final temperature of the reaction is:

22.71° C

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Answer:

Check Explanation

Explanation:

a) The balanced chemical reaction between Aluminium and Silver Sulfide is represented below

2Al + 3Ag₂S → Al₂S₃ + 6Ag

Aluminium displaces Silver from Silver sulfide because it is higher than Silver in the electrochemical series.

b) Names of the products from this reaction

Ag is called Silver metal. Free Silver metal.

Al₂S₃ is called Aluminium Sulfide.

c) This reaction is a single-displacement reaction because an element directly displaces and replaces another element in a compound.

It is also a redox reaction (reduction-oxidation reaction) because there are species being oxidized and reduced simultaneously!

d) Yes, this reaction is an oxidation-reduction (redox) reaction because there are species being oxidized and reduced simultaneously!

e) The specie that is being oxidized is said to undergo oxidation. And oxidation is defined as the loss of electrons, thereby leading to an increase in oxidation number.

In this reaction, it is evident that Aluminium undergoes oxidation as its oxidation number increases from 0 in the free state to +3 when it displaces Silver and becomes Aluminium sulfide.

Al → Al³⁺

0 → +3 (Oxidation)

The specie that is being reduced is said to undergo reduction. And reduction is defined as the gain of electrons, thereby leading to a decrease in oxidation number.

In this reaction, it is evident that the silver ion undergoes reduction as its oxidation number decreases from +1 in the Silver Sulfide compound to 0 when it is displaced and becomes Silver in free state.

Ag⁺ → Ag

+1 → 0 (Reduction)

The reducing agent is the specie that brings about reduction. Since Silver ion in Silver sulfide is reduced, Aluminium is the reducing agent that initiates the reduction process.

The oxidation agent is the specie that brings about oxidation. Since, Aluminium is the specie that undergoes oxidation, the oxidizing agent is the Silver Sulfide that brings about the oxidation.

Hope this Helps!!!

The new volume of the Cl2 gas under the conditions at the bottom of Hellas Planitia on Mars would be approximately 4.2 litres, as calculated using the ideal gas law.

The question is asking about the adjustment of gas volume under different conditions of pressure and temperature, so the ideal gas law is applicable here which states that: PV = nRT, where P is pressure, V is volume, T is temperature, n is number of moles of the gas and R is gas constant.

At Standard Temperature and Pressure (STP), the conditions are 1 atm pressure and 0°C (273 K). Please note that the volume given is already at STP, so to find the new volume, we should rearrange the ideal gas law to V2 = V1 * P1/P2 * T2/T1.

However, note that all the measurements need to be in the same standard units. Here, the initial pressure (P1) is 1 atm, converted to kPa becomes approximately 101.3 kPa, the final pressure (P2) is given as 1.16 kPa. The initial temperature (T1) is 273K and the final temperature (T2) needs to be converted from Celsius to Kelvin, making it 268 K (-5°C + 273). Thus, applying the equation, we get:

V2 = 0.175L * 101.3 kPa / 1.16 kPa * 268 K / 273 K = 4.2 L approximately

Therefore, the new volume of the Cl2 gas, if it is transported to the conditions at the bottom of Hellas Planitia on Mars, would be approximately 4.2 litres.

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Using the combined gas law, the new volume of Cl₂ gas at 1.16 kPa pressure and -5.0°C temperature at the bottom of Hellas Planitia on Mars is calculated to be approximately 15.06 liters.

To determine the new volume of Cl₂ gas at the conditions provided, we need to use the combined gas law which is given by the equation: [tex]\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}[/tex], where P is pressure, V is volume and T is temperature in Kelvin.

At STP (Standard Temperature and Pressure), we have a volume of 175 mL (or 0.175 liters), a pressure of 101.3 kPa, and a temperature of 273.15 K. The conditions at the bottom of the Hellas Planitia are 1.16 kPa and -5.0°C or 268.15 K.

First, we convert the initial volume into liters: 175 mL = 0.175 L.
Next, we rearrange the combined gas law to solve for V₂:

Therefore, the new volume of Cl₂ gas at the bottom of Hellas Planitia will be approximately 15.06 liters.

Answer:

13.279

Explanation:

24g/56.1056g/mol= 0.42 (mol KOH)

0.42 mol KOH/ 2.25 L= 0.19 M KOH

-log(0.19)= pOH= 0.72

pH=14-pOH

pH=14-0.72=13.279

sorry it's not the best explanation but since you're on a time crunch i hope this helps

A chemist prepared a solution of KOH by completely dissolving 24.0 grams of solid KOH in 2.25 liters of water at room temperature. The pH of the solution that the chemist prepared, to the nearest thousandth is 13.279.

pH is a numerical indicator of how acidic or basic aqueous or some other liquid solutions are. The phrase, which is frequently used during chemistry, biology, or agronomy, converts hydrogen ion concentrations.

The hydrogen ion concentration in pure water, which has a pH of 7, is 107 gram-equivalents per liter, making it neutral.

24g/56.1056g/mol= 0.42 (mol KOH)

0.42 mol KOH/ 2.25 L= 0.19 M KOH

-log(0.19)= pOH= 0.72

pH=14-pOH

pH=14-0.72

   =13.279

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Answer:

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Explanation:

[tex]\Delta T_f=T-T_f[/tex]

[tex]\Delta T_f=K_f\times m[/tex]

[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]

where,

[tex]T[/tex] = Freezing point of solvent

[tex]T_f[/tex] = Freezing point of solution

[tex]\Delta T_f[/tex] =depression in freezing point

[tex]K_f[/tex] = freezing point constant

m = molality

we have :

Mass of ethylene glycol = 59.0 g

Molar mass of ethylene glycol = 62.1 g/mol

Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)

[tex]K_f[/tex] =1.86°C/m ,

[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]

[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]

[tex]\Delta T_f=3.26^oC[/tex]

Freezing point of pure water = T =  0°C

Freezing point of solution = [tex]T_f[/tex]

[tex]\Delta T_f=T-T_f[/tex]

[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]

The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.

Answer:

c. 3.00 M HCl

Explanation:

From dilution formula

C1V1 = C2V2

C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml

Substitute and Simplify

C1×10= 1.5×20

C1= 3.00M

Answer:

The empirical formula of the compound is [tex]CrBr_2[/tex].

Explanation:

Mass of chromium = 19.13 g

Mass metal bromide formed = 77.92 g

Mass of bromine in metal bromide = 77.92 g - 19.13 g = 58.79 g

Moles of chromium metal :

[tex]=\frac{19.13 g}{52 g/mol}=0.3679 mol[/tex]

Moles of bromine:

[tex]=\frac{58.79 g}{80 g/mol}=0.7349 mol[/tex]

For empirical formula divide the smallest number of moles of element from the all the moles of elements:

chromium : [tex]\frac{0.3679 mol}{0.3679}=1[/tex]

bromine : [tex]\frac{0.7349 mol}{0.3679}=2[/tex]

The empirical formula of the compound is [tex]CrBr_2[/tex].

To find the empirical formula, the moles of each element involved must be determined. When the sample weights are converted to moles, the ratio of chromium (Cr) to bromine (Br) in the bromide compound is found to be 1:2, making the empirical formula CrBr2.

To determine the empirical formula of the compound, we first need to identify the moles of each element. The chromium sample weighs 19.13g. Using the atomic weight of chromium, which is 52.00g/mol, the number of moles of chromium is 19.13g / 52.00g/mol = 0.368 mol.

Considering the entire mass of the metal bromide, which is 77.92 g, the mass of bromine in the compound must be 77.92 g - 19.13 g = 58.79 g. As the atomic weight of bromine is approximately 79.90g/mol, we can find that the moles of bromine are 58.79 g / 79.90 g/mol = 0.736 mol.

The ratio of Cr to Br in the compound is 0.368 : 0.736, which is approximately 1 : 2 when we divide both numbers by the smallest to obtain the simplest whole number ratio. Therefore, the empirical formula of the compound is CrBr2.

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Final answer:

To find the molar mass of the diprotic acid, divide the mass of the sample by the number of moles. The molar mass of the acid is 307.03 g/mol.

Explanation:

To find the molar mass of the diprotic acid, we can use the information from the titration. First, we need to determine the number of moles of NaOH used in the titration. The volume of NaOH solution used is 35.8 mL, which is equal to 0.0358 L. The molarity of the NaOH solution is 0.160 M.

Therefore, the number of moles of NaOH used is 0.0358 L x 0.160 mol/L = 0.005728 mol.

Since the diprotic acid is diprotic, it can donate two moles of H+ ions per mole of acid. Therefore, the number of moles of the diprotic acid is half of the number of moles of NaOH used, which is 0.005728 mol / 2 = 0.002864 mol.

To calculate the molar mass of the acid, we need to divide the mass of the sample by the number of moles. The mass of the acid sample is 0.881 g. Therefore, the molar mass of the acid is 0.881 g / 0.002864 mol = 307.03 g/mol.