The properties of elements are different than the compound that the elements form. Is this statement True or False

Answer:

molar mass (of the diprotic acid) = 116g/mol

[tex]pK_{a_1}= 1.83[/tex]

[tex]pK_{a_2}=6.07[/tex]

Explanation:

The number of moles of [tex]NaOH[/tex] and [tex]H^+[/tex] are equal at the equivalence point since they are both taking part in the diprotic acid.

0.1 M means 0.1 moles in 1L or 0.1 moles in 1000 mL

Number of moles of 0.1 M NaOH at final equivalence point ;

=  [tex]34.72 mL * \frac{0.1mole}{1000mL}[/tex]

= 0.00347 moles of NaOH

However, the number of moles of the diprotic acid in the 0.25 L solution is = [tex]\frac {0.00347}{2 }[/tex]      (due to the fact that half of the concentration of NaOH is needed to give the same amount of  [tex]H^+[/tex]

[tex]= 0.001736[/tex]  

Given that:

The  Total acid in the solution = 0.2015 g; to calculate the molar mass ; we have :

no of moles = [tex]\frac{mass}{molar mass}[/tex]

molar mass= [tex]\frac{mass}{numbers of moles}[/tex]

molar mass = [tex]\frac{0.2015}{0.001736}[/tex]

molar mass (of the diprotic acid) = 116g/mol

At the second equivalence point;

The pH = 9.27

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

[tex][H^+]= 10^{-pH}[/tex]

[tex][H^+] = 10^{-9.27[/tex]

[tex][H+] = 5.37 * 10^{-10[/tex]

[tex][OH^-][/tex] can be calculated as follows:

[tex]\frac{10^{-14}}{[H^+]} = \frac {10^{-14}}{5.37* 10^{-10}}[/tex]

[tex][OH^-][/tex] = [tex]1.86*10^{-5}[/tex]

Let represent the equation from the reaction after the second equivalence point with:

[tex]X^{2-} + H_2O \rightleftharpoons HX^- + OH^-[/tex]

where:

[tex]X^{2-}[/tex] =  [tex]\frac{0.001736}{(25 + 34.72)} * 1000[/tex]

[tex]X^{2-}[/tex] =  [tex]0.029 M[/tex]

 The ICE Table is shown as follows;

                      [tex]X^{2-}[/tex]    [tex]+[/tex]    [tex]H_2O[/tex]       [tex]\rightleftharpoons[/tex]      [tex]HX^-[/tex]     [tex]+[/tex]        [tex]OH^-[/tex]

Initial            0.029                                  0                     0

Change        -x                                         +x                   +x  

Equilibrium  (0.029 - x)                            x                     x

[tex]Kb = \frac{[HX^-][OH^-]}{[X^{2-}]}[/tex]

[tex]Kb = \frac{x^2}{0.029-x}[/tex]

[tex]k_b}= \frac{(1.86*10^{-5})^2}{0.029-(1.86*10^{-5})}[/tex]

[tex]x = 1.191*10^{-8}[/tex]

[tex]K_a} = \frac {10^{-14}}{K_b}[/tex]

[tex]K_a} = \frac {10^{-14}}{1.191*10^{-8}}[/tex]

[tex]K_a} = 8.39 * 10^{-7}[/tex]

[tex]pK_a} = -log K_a[/tex]

[tex]pK_a} = -log (8.39*10^{-7})[/tex]

[tex]pK_{a_2}=6.07[/tex]

At the first equivalence point we have all H2X getting converted to  [tex]HX^-[/tex]  [tex]HX^-[/tex] is an amphoteric species which implies that it can serve as both an acid and a base

As such, in this process:

[tex]pH = pK_{a_1} + \frac{pKa_2}{2}[/tex]

Given that: the pH = 3.95

Then;

[tex]3.95 = pK_{a_1} + \frac{6.07}{2}[/tex]

[tex]3.95*2 = pK_{a_1} +{6.07}[/tex]

[tex]7.9 = pK_{a_1} +{6.07}[/tex]

[tex]pK_{a_1}= 7.9 - {6.07}[/tex]

[tex]pK_{a_1}= 1.83[/tex]

Answer:

gfhgfhhgjffhfhgfhgfhfgfhfg

Explanation:

Answer:

7.60 M

Explanation:

Our method to solve this question is to  use the definition of molarity (M) concentration which is the number of moles per liter of solution, so for this problem we have

[Cl⁻] = # mol Cl⁻ / Vol

Now the number of moles of Cl⁻ will be sum of Cl in the 1.00 mL 5.4 M solution plus the moles of Cl⁻ in the 0.50 mL 12 M H . Since the volume in liters times the molarity gives us the number of moles we will have previous conversion of volume to liters for units consistency:

1mL x 1 L / 1000 mL = 0.001 L

0.5 mL x 1L/1000 mL = 0.0005 L

[Cl⁻]  =  0.001 L x 5.4 mol/L + 0.0005L x 12 mol/L / ( 0.001 L + 00005 L )

= 7.6 M

This is the same as the statement given in the question.

Answer:

We need 7.55 grams of PCl3

Explanation:

Step 1: Data given

Mass of Cl2 = 3.90 grams

Mass of PCl5 = 11.45 grams

Molar mass Cl2 = 70.9 g/mol

Molar mass PCl5 = 208.24 g/mol

Step 2: The balanced equation

PCl3 + Cl2 → PCl5

Step 3: calculate moles

Moles = mass / molar mass

Moles Cl2 = 3.90 grams / 70.9 g/mol

Moles Cl2 = 0.0550 moles

Moles PCl5 = 11.45 grams /208.24 g/mol

Moles PCl5 = 0.0550 moles

Step 4: Calculate moles PCl3

For 1 mol PCl3 we need 1 mol Cl2 to produce 1 mol PCl5

For 0.0550 moles PCl5 we will need 0.0550 moles Cl2 and 0.0550 moles PCl3

Step 5: Calculate mass PCl3

Mass PCl3 = moles * molar mass

Mass PCl3 = 0.0550 moles * 137.33 g/mol

Mass PCl3 = 7.55 grams

Without calculating the number of moles:

Mass of Cl2 + mass PCl3 = mass PCl5

Mass PCl3 = 11.45 - 3.90 = 7.55 grams

We need 7.55 grams of PCl3

Final answer:

To find the mass of PCl3 needed for the reaction, use the mass difference between the product PCl5 and the reactant Cl2, resulting in 7.55 g of PCl3.

Explanation:

The question involves a stoichiometry calculation to determine the amount of phosphorus trichloride (PCl₃) required to react with chlorine gas (Cl₂) to produce phosporus pentachloride (PCl₅). If 3.90 g of Cl₂ reacts to form 11.45 g of PCl₅, we can use the principle of conservation of mass to find the mass of PCl₃ that reacted. According to the reaction PCl₃ + Cl₂
PCl₅, for every mole of PCl₅ produced, one mole of PCl₃ is consumed.

In this problem, since we start with Cl₂ and end up with PCl₅, the difference in mass between PCl₅ and Cl₂ must be the mass of PCl₃ consumed. Therefore, the mass of PCl₃ is given by subtracting the mass of Cl₂ from the mass of PCl₅:

Mass of PCl₃ = Mass of PCl₅ - Mass of Cl₂
= 11.45 g - 3.90 g
= 7.55 g

Thus, 7.55 g of PCl₃ is needed for the reaction.

The question is incomplete, here is the complete question:

To analyze the experiment used to determine the properties of an electron. In 1909, Robert Millikan performed an experiment involving tiny, charged drops of oil. The drops were charged because they had picked up extra electrons. Millikan was able to measure the charge on each drop in coulombs. Here is an example of what his data may have looked like.

 Drop        Charge (C)

 A             -3.20 × 10⁻¹⁹

 B             -4.80 × 10⁻¹⁹

 C             -8.00 × 10⁻¹⁹

 D             -9.60 × 10⁻¹⁹

Based on the given data, how many extra electrons did drop C contain? Express your answer as an integer.

Answer: The extra electrons that the drop C contain are 5

Explanation:

Millikan’s oil drop experiment is used to measure the charge of an electron. Before this experiment, the subatomic particles were not accepted.

He found that all the oil drops had charges that were the multiples of [tex]-1.6\times 10^{-19}C[/tex]. This value is the charge on 1 electron

Number of electrons excess electrons is calculated by using the formula:

[tex]\text{Excess electrons}=\frac{\text{Charge on millikan's oil drop}}{\text{Charge on 1 electron}}[/tex]

For Drop C:

Charge on drop C = [tex]-8.00\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-8.00\times 10^{-19}}{-1.6\times 10^{-19}}=5[/tex]

Hence, the extra electrons that the drop C contain are 5

Question: complete the following sentences: aldohexose, triose,aldose, ketopentose, glucose, ketose.

1) Glyceraldehyde is an example of a(n)____ because it has three carbon atoms.2) With the carbonyl group on the end of a six-month carbon chain, the carbohydrate would be classified as a( n)__3)Any carbohydrate with the carbonyl group on the second carbon is a(n)___4) A monosaccharide is a(n)___ if the carbonyl group is on the end of the carbon chain. 5)The most common carbohydrate ___ has six atoms. 6) If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n)_____.

Answer:

1) triose

2) aldohexose

3) ketose

4) aldose

5) glucose

6) ketopentose

Explanation:

Carbohydrate is one of the major classes of food which is divided into monosaccharide, disaccharide and polysaccharide. Monosaccharide is the simplest form of carbohydrates which is further divided into:

- glucose( the most common carbohydrate)

- fructose and

- galactose.

These monosaccharides can be further grouped according to the number of carbon atoms they possess and according to the functional group attached to its linear or unbranched carbon skeleton.

-TRIOSE: These are monosaccharides that has three carbon atoms example is Glyceraldehyde.

-ALDOHEXOSE: These are monosaccharides with the carbonyl group on the end of a six carbon chain.

- KETOSE: These are monosaccharides with the carbonyl group on the second carbon.

- ALDOSE: This is a monosaccharide in which the carbonyl group is on the end of the carbon chain.

-KETOPENTOSE: These are monosaccharides that has five carbon atoms and a carbonyl group on the second carbon. An example is xylulose.

This question is about the classification of carbohydrates based on the position of the carbonyl group.

1. Glyceraldehyde is an example of an aldose because it contains an aldehyde functional group and it has three carbon atoms.
2. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as an aldohexose.
3. Any carbohydrate with the carbonyl group on the second carbon is classified as a ketose.
4. A monosaccharide is categorized as an aldotriose if the carbonyl group is on the end of the carbon chain.
5. The most common carbohydrate, glucose, has six carbon atoms and is an example of an aldohexose.
6. Xylulose is a ketopentose because it has five carbon atoms and a carbonyl group on the second carbon.

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The balanced equation is

2 C6H14 (g) + 19 O2 (g) --> 12 CO2 (g) + 14 H2O (g)

For solving the stoichiometric calculations, we first need to do two steps. One of them is balancing the reaction equation. Here the balancing is done and we can see that 2 moles of hexane reacts with 19 moles of oxygen to produce 12 moles of carbon dioxide and 14 moles of water.

Now, for 2 moles of hexane, number of moles of oxygen required is 19.

So, for 7.2 moles of hexane, number of moles of oxygen required is[tex]\frac {19}{2}\times 7.2[/tex].

= 68.4.

So 68.4 moles of oxygen is required.

68.4 moles of oxygen is required.

Balanced chemical equation:

2 C₆H₁₄ (g) + 19 O₂ (g) ------> 12 CO₂ (g) + 14 H₂O (g)

2 moles of hexane reacts with 19 moles of oxygen to produce 12 moles of carbon dioxide and 14 moles of water.

Now, for 2 moles of hexane, number of moles of oxygen required is 19.

So, for 7.2 moles of hexane, number of moles of oxygen required = 68.4.

So, 68.4 moles of oxygen is required.

Find more information about Mole-ratio here:

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Answer:

Weak acid - strong base

Explanation:

H₃PO₄ → Phosphoric acid.

This is a weak that has three dissociations in order to give hydronium to the medium and to produce the phosphate anion. The equations are:

H₃PO₄  + H₂O  ⇄  H₃O⁺  +  H₂PO₄⁻               Ka1

H₂PO₄⁻  + H₂O  ⇄  H₃O⁺ +  HPO₄⁻²               Ka2

HPO₄⁻²  + H₂O  ⇄  H₃O⁺  +  PO₄⁻₃             Ka3

As the H₃PO₄ is a weak acid then the water behaves as a strong base.

If we follow the Brownsted Lory idea, water becomes a strong base cause it receives the H⁺ from water, then the H₃O⁺ becomes the conjugate weak acid.

Anions from the H₃PO₄,  diacid phosphate and monoacid phosphate assume the rol of the conjugate strong base, they all recieve proton but this is a special case, because both anions can recieve all release the proton. That's why, they also are amphoteric

Answer:

We need 50.6 grams of oxygen

Explanation:

Step 1: Data given

Mass of ethane = 13.6 grams

Molar mass of ethane = 30.07 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 13.6 grams / 30.07 g/mol

Moles ethane = 0.452 moles

Step 4: Calculate moles oxygen

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.452 moles ethane we need 3.5*0.452 = 1.582 moles O2

Step 5: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 1.582 moles * 32.0 g/mol

Mass O2 =  50.6 grams

We need 50.6 grams of oxygen

The balanced chemical equation for the complete combustion of ethane (C2H6) in oxygen (O2) is: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O. To burn 13.6g of ethane completely, approximately 50.63g of oxygen is needed.

Firstly, the balanced chemical equation for the complete combustion of ethane (C2H6) in oxygen (O2) is:

C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O

From this equation, one mole of ethane reacts with 3.5 moles of oxygen. To calculate the mass of oxygen needed, you firstly need to know the molar mass of ethane (C2H6), which is approximately 30.07 g/mol. Therefore, 13.6 g of ethane is about 0.452 moles.

Since 1 mole of ethane reacts with 3.5 moles of oxygen, then 0.452 moles of ethane would require (0.452 x 3.5) = 1.582 moles of oxygen. The molar mass of oxygen (O2) is about 32 g/mol, so the mass of oxygen required is (1.582 moles x 32 g/mol) = 50.63 g.

Therefore, the mass of oxygen required for the complete combustion of 13.6 g of ethane is approximately 50.63 grams.

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Answer:

[tex]C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}[/tex]

Explanation:

The differential equation is given as:

[tex]\frac{dC}{dt} = r- kC[/tex]

[tex]\frac{dC}{r- kC} = dt[/tex]

Taking integral of both sides; we have:

[tex]\int\limits \frac{dC}{r- kC} = \int\limits dt[/tex]

[tex]-\frac{1}{k} In(r-kC) = t+D\\In(r-kC)=-kt-kD[/tex]

[tex]r-kC=e^{-kt-kD}[/tex]

[tex]r-kC=e^{-kt}e^{-kD}[/tex]

[tex]r-kC=Ae^{-kt}[/tex]

[tex]kC=r-Ae^{-kt}[/tex]

[tex]C=\frac{r}{k}-\frac{A}{k}e^{-kt}[/tex]

[tex]C(t)=\frac{r}{k}-\frac{A}{k}e^{-kt}[/tex]     ------- equation (1)

If C(0)= [tex]C_o[/tex] ; we have:

C(0)= [tex]\frac{r}{k}-\frac{A}{k}e^0[/tex]         (where; A is an integration constant)

[tex]C_o = \frac{r}{k}- \frac{A}{k}[/tex]

[tex]C_o=\frac{r-A}{k}[/tex]

[tex]kC_o=r-A[/tex]

[tex]A=r-kC_o[/tex]

Substituting [tex]A=r-kC_o[/tex] into equation (1); we have;

[tex]C(t)=\frac{r}{k}-(\frac{r-kC_o}{k})e^{-kt}[/tex]

Answer:

Theoretical yield for CO₂ is 5.10g

Explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g)  + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g

Answer:

The theoretical yield of carbon dioxdide = 5.11 grams

Explanation:

Step 1: Data given

Mass of oxygen gas = 4.64 grams

Molar mass of O2 = 32.0 g/mol

The reaction yields 3.95 grams of carbon dioxide (CO2)

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

2C6H6 + 15O2 → 12CO2 + 6H2O

Step 3: Calculate moles oxygen

Moles oxygen = mass oxygen / molar mass oxygen

Moles oxygen = 4.64 grams / 32.0 g/mol

Moles oxygen = 0.145 moles

Step 4: Calculate moles of carbon dioxide (CO2)

For 2 moles C6H6 we need 15 moles O2 to produce 12 moles CO2 and 6 moles H2O

For 0.145 moles O2 we'll have 12/15 * 0.145 = 0.116 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.116 moles * 44.01 g/mol

Mass CO2 = 5.11 grams

The theoretical yield of carbon dioxdide = 5.11 grams

The question is incomplete, here is the complete question:

1. The radioactive source you will be working with in this lab is Cs-137. Look up the half-life of this material and report the value in units of seconds.

2. The relationship between decay constant (l) and half-life is:

[tex]t_{1/2}=\frac{\ln 2}{k}[/tex]

For Cs-137, what is the value of 'k' in [tex]s^{-1}[/tex]

Answer:

For 1: The half life for Cs-137 isotope is [tex]9.51\times 10^8s[/tex]

For 2: The rate constant of Cs-137 isotope is [tex]7.29\times 10^{-10}s^{-1}[/tex]

Explanation:

Half life is defined as the time taken for half of the reaction to complete. This is also defined as the time in which the concentration of a reactant is reduced to half of its original value.

The half life for Cs-137 isotope is [tex]9.51\times 10^8s[/tex]

The relationship between decay constant (l) and half-life is: given by the equation:

[tex]t_{1/2}=\frac{\ln 2}{k}[/tex]

where,

[tex]t_{1/2}[/tex] = half life of Cs-137 isotope = [tex]9.51\times 10^8s[/tex]

k = rate constant

Putting values in above equation, we get:

[tex]9.51\times 10^8s=\frac{\ln 2}{k}\\\\k=\frac{\ln 2}{9.51\times 10^8}=7.29\times 10^{-10}s^{-1}[/tex]

Hence, the rate constant of Cs-137 isotope is [tex]7.29\times 10^{-10}s^{-1}[/tex]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

Answer:

(a) 13.69

(b) i beaker 1: 0g

    ii beaker 2: 0g

Explanation:

a. The solubility equilibrium equation for KCl is

KCl(s)  ⇄  K⁺(aq)  +  Cl⁻(aq)

3.7M KCl contains equal moles of K ions and Cl ions

therefore, the ion-product expression is written thus

Ksp = [K⁺][Cl⁻]

       = [3.7][3.7]

       = 13.69

b. from the first two beakers containing 100 mL and 3.7M KCl

moles of K⁺ = moles of Cl⁻ = moles of KCl = 3.7moles in 1L

if 3.7M Implies 3.7 moles in 1L or 1000 mL or 1000 cm³

how many moles will be contained in 100 mL

this is calculated as follows

3.7moles/Liter * 100 mL

[tex]\frac{3.7 moles KCl}{1000 mL} * 100 ml = 0.37moles KCl[/tex]

= 0.37moles K⁺ = 0.37moles Cl⁻

4.0 M HCl, contains

[tex]\frac{4 moles HCl}{1000 mL} *100mL = 0.4moles HCl = 0.4 moles H = 0.4moles Cl[/tex] in 100mL

8.0M HCl, contains

[tex]\frac{8moles HCL}{1000mL} *100mL=0.8mole HCl=0.8molesH=0.8molesCl[/tex] in 100mL

now, in the first beaker 100 mL of 4M HCl is added to 100 mL of 3.7M KCl

total moles of Cl⁻ (0.4 + 0.37) moles = 0.77 moles

total moles of K⁺ remains 0.37 moles

total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L

total moles of Cl⁻ per Liter = 0.77moles/0.2L = 3.85M Cl⁻

total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺

Qsp must be greater or equal to Ksp for Precipitation to occur, that is

Qsp ≥ Ksp

Qsp = [K][Cl] = [1.85][3.85] = 7.12 this is less than 13.69(Ksp)

hence no KCl will precipitate in the first beaker

since there is no precipitate, there is therefore no need for calculating the mass precipitated

and the answer is 0g

(bii) now, in the second beaker 100 mL of 8M HCl is added to 100 mL of 3.7M KCl

total moles of Cl⁻ (0.8 + 0.37) moles = 1.17 moles

total moles of K⁺ remains 0.37 moles

total volume of solution = (100mL + 100mL) = 200mL/1000mL = 0.2L

total moles of Cl⁻ per Liter = 1.17moles/0.2L = 5.85M Cl⁻

total moles of K⁺ per Liter = 0.37moles/0.2L = 1.85M K⁺

Qsp must be greater or equal to Ksp for Precipitation to occur, that is

Qsp ≥ Ksp

Qsp = [K][Cl] = [1.85][5.85] = 10.82 this is less than 13.69(Ksp)

hence no KCl will precipitate also in the second beaker

since there is no precipitate, there is therefore no need fo calculating the mass precipitated

and the answer is 0g

The monatomic ion with a charge of +2 and electronic configuration 1s22s22p63s23p6 is isoelectronic with the noble gas argon (Ar). The chemical symbol of the ion is Ca^2+.

The given electronic configuration is 1s22s22p63s23p6. This indicates that it has a total of 18 electrons, which is the same as the electronic configuration of the noble gas argon (Ar). Therefore, the monatomic ion is isoelectronic with argon.

The charge of the ion is +2, which means it has lost 2 electrons from its neutral state. Since it has 18 electrons, it must have had 20 electrons in its neutral state. The symbol of the ion can be determined by looking at the element that has an atomic number of 20, which is calcium (Ca). Therefore, the chemical symbol of the ion is Ca2+.

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Answer: Option (A) is the correct answer.

Explanation:

Exothermic reaction is defined as the reaction in which release of heat takes place. Whereas endothermic reaction is defined as the reaction in which heat is absorbed by the reactant molecules.

Thus, we can conclude that for the given statements its is true that ALL three of the given processes are endothermic.

Answer: D) forming mixtures from pure substances.

Explanation

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

a) Increase in temperature of a gas: As increasing the temperature , increases the kinetic energy of molecules , the molecules move faster and thus entropy increases.

b) Melting a solid : The randomness will increase as liquids move freely as compared to solids and hence entropy will also increase.

c) chemical reactions that increase the number of moles of gas: As more gaseous molecules will be formed, more will be the randomness and hence entropy increases.

d) forming mixtures from pure substances: As substances in a mixture do not react chemically and thus the molecules remain same and entropy remain same.

e) decreasing the volume of a gas: According to Boyle's law, decreasing the volume will increase the pressure and thus entropy will increase.

Final answer:

The process that does not lead to an increase in entropy is decreasing the volume of a gas as it compact the molecules into a smaller space, reducing disorder.

Explanation:

The question revolves around identifying which process does not lead to an increase in entropy. Entropy is a measure of the disorder or randomness in a system. Several processes can increase entropy, such as:

Increasing the temperature of a gas, which increases the kinetic energy and disorder of the gas molecules.

The process of melting a solid, where the ordered solid structure becomes a more disorderly liquid.

Chemical reactions that increase the number of gas molecules, as a greater number of particles usually means greater disorder.

Forming mixtures from pure substances, whereby the uniform structure becomes more randomly arranged upon mixing.

However, the one process that does not increase entropy is decreasing the volume of gas. Decreasing the volume of a gas reduces its entropy because it compacts the gas molecules into a smaller space, potentially giving the system less disorder.

Therefore, the correct answer is E) decreasing the volume of a gas.

Answer:

D) n-hexane<2-methylpentane <2, 3-dimethylbutane

Explanation:

structures with less branching have  high boiling point and low melting points this is because of freezing depression and boiling depression and stronger bonds.

Answer:

True

Explanation:

The gaseous state is characterized in that the cohesion forces are usually null, in which the particles have their maximum mobility. The particles tend to occupy all the available volume, so their shape and volume are variable. The gaseous state is a dispersed state of matter, which means that the molecules are separated by distances much larger than the diameter of the gas molecules.

True, a gas is a substance where the molecules are relatively far apart compared to their size. Gases can diffuse until they evenly fill their container, unlike solids and liquids. For example, oxygen we breathe spreads throughout a room.

True, a gas is a substance in which the molecules are separated on the average by distances that are large compared with the sizes of the molecules. This is because gases have the capability to diffuse and spread out until they evenly fill their container. This differentiates them from solids and liquids, where the intermolecular distances are much smaller. For instance, the oxygen we breathe spreads throughout a room, as opposed to settling in one place.

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Answer: The standard potential of the cell is 0.77 V

Explanation:

We know that:

[tex]E^o_{Ni^{2+}/Ni}=-0.25V\\E^o_{Cu^{+}/Cu}=0.52V[/tex]

The substance having highest positive [tex]E^o[/tex] reduction potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction: [tex]Ni(s)\rightarrow Ni^{2+}(aq)+2e^-[/tex]

Reduction half reaction: [tex]Cu^{+}(aq)+e^-\rightarrow Cu(s)[/tex]       ( × 2)

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Putting values in above equation follows:

[tex]E^o_{cell}=0.52-(-0.25)=0.77V[/tex]

Hence, the standard potential of the cell is 0.77 V

Answer:

The rate law is k = [tex]k\frac{[O3]^{2} }{[O2]}\\[/tex]

Explanation:

From the mechanism is necessary to derive the rate law from the elementary steps and explain the effects of [O2] on the rate

The first step is a reversible reaction. Assuming dynamic equilibrium is achieved, the rate of the forward reaction is equal to the rate of the backward reaction

rate(forward) = rate(backward)

k1 [O3] = k-1 [O] [O2]

[O] is not part of the rate law, so we need to express [O] in terms of other species

[O] = [tex]\frac{k1 [O3]}{k-1[O2]}[/tex]

from the second step

rate = k2[O] [O3]

substituting [O] from the first step

rate = [tex]k2 \frac{k1 [O3] [O3]}{k-1[O2]} = \frac{k2k1 [O3]^{2} }{k-1[O2]}\\[/tex]

k = [tex]\frac{k2k1}{k-1}\\[/tex]

The final rate law is then

k = [tex]k\frac{[O3]^{2} }{[O2]}\\[/tex]

So, as the concentration os O2 increase the rate decrease. Also from the first step of the mechanism we can se that O2 can react to O to form back the reactant O3 resulting in decreased reaction rate.

Final answer:

The rate law for the decomposition of ozone in terms of the constants k1, k2, and k−1 is rate = k1k2[O3]^2 / (k−1 + k2[O3]). This is derived using the proposed two-step mechanism and making an assumption that the backward reaction is much slower than the forward reaction.

Explanation:

The proposed mechanism for the decomposition of ozone to molecular oxygen involves two elementary steps:

O3 O + O2 (rate constant k1)

O + O3 2O2 (rate constant k2)

There is also a reverse reaction to step 1 that should be considered, where O and O2 combine to reform O3 (rate constant k⁻1). According to the steady-state approximation, the concentration of intermediate species O remains constant because it is produced and consumed at the same rate. Therefore, the rate of its formation in step 1 is equal to its rate of consumption in step 2 and any reverse reaction.

The rate law for the given mechanism can be expressed as:

rate = k2[O][O3]

However, [O] is not directly measurable, so we need to link it to other reactants. We can express [O] from the equilibrium of step 1, assuming that the backward reaction is much slower than the forward reaction:

[O] = k1[O3] / (k⁻1 + k2[O3])

Then by substituting [O] back into the rate law:

rate = k2(k1[O3] / (k⁻1 + k2[O3]))[O3]

This simplifies to:

rate = k1k2[O3]^2 / (k⁻1 + k2[O3])

This rate law now relates the overall rate to the experimental rate constants k1, k2, and k⁻1.

Answer:

9.96g of Na₂SO₄ is the maximum mass that could be produced

Explanation:

We must determine the reaction:

Reactants: H₂SO₄, NaOH

Products: H₂O, Na₂SO₄

The equation is:  H₂SO₄ (aq) + 2NaOH(aq) →  2H₂O (l) + Na₂SO₄(aq)

We have the mass of the reactants. We need to convert them to moles, in order to define the limiting reactant

6.87 g / 98 g/mol = 0.0701 moles of acid

9.7 g / 40 g/mol = 0.242 moles of base

Limiting reactant is the acid. Let's verify

2 moles of NaOH can react with 1 mol of acid

Therefore 0.242 moles of NaOH must react with (0.242 . 1) / 2 = 0.121 moles

We do not have enough acid.

Ratio with the salt is 1:1. 1 mol of acid produces 1 mol of salt

Therefore 0.0701 moles of acid will produce 0.0701 moles of salt

We convert the moles to mass → 0.0701 mol . 142.06 g / 1 mol = 9.96g

The heat of vaporization, ΔHvap, of pinene is 41094 Joules.

Here the following formula should be used.

ln(P2/P1) = ΔHvap/R (1/T1 - 1/T2)

Here,

P1 = initial pressure at  429 K = 760 torr

P2 = final pressure at 415 K  = 515 torr

R = gas constant = 8.314 J/mole.K

T1 = initial temperature =  429 K

T2 = final temperature = 515 K

So, the heat should be

log(515/760) =  ΔH/2.303*8.314 {1/429k - 1/415k)

= 41094 J

Hence, The heat of vaporization, ΔHvap, of pinene is 41094 Joules.

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The heat of vaporization (ΔHvap) of pinene is calculated to be approximately 40.3 KJ/mol.

To calculate the heat of vaporization [tex](\(\Delta H_{\text{vap}}\))[/tex] of pinene, we can use the Clausius-Clapeyron equation, which relates the vapor pressure and temperature to the heat of vaporization:

[tex]\[\ln \left( \frac{P_1}{P_2} \right) = \frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)\][/tex]

Where:

- [tex]\( P_1 \)[/tex] and [tex]\( P_2 \)[/tex] are the vapor pressures at temperatures [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex], respectively.

- R is the universal gas constant (8.314 J/mol·K).

- [tex]\( \Delta H_{\text{vap}} \)[/tex] is the heat of vaporization.

Given data:

- [tex]\( P_1 = 760 \)[/tex] torr at [tex]\( T_1 = 429 \)[/tex] K

- [tex]\( P_2 = 515 \)[/tex] torr at [tex]\( T_2 = 415 \)[/tex] K

First, we convert the pressures to the same units if necessary, but here they are both in torr.

Now, let's use the Clausius-Clapeyron equation to solve for [tex]\(\Delta H_{\text{vap}}\)[/tex]:

[tex]\[\ln \left( \frac{760}{515} \right) = \frac{\Delta H_{\text{vap}}}{8.314} \left( \frac{1}{415} - \frac{1}{429} \right)\][/tex]

Calculate the natural logarithm and the reciprocal temperatures:

[tex]\[\ln \left( \frac{760}{515} \right) = \ln(1.4757) \approx 0.388\][/tex]

[tex]\[\frac{1}{415} \approx 0.00241 \quad \text{K}^{-1}\][/tex]

[tex]\[\frac{1}{429} \approx 0.00233 \quad \text{K}^{-1}\][/tex]

[tex]\[\frac{1}{415} - \frac{1}{429} \approx 0.00241 - 0.00233 = 0.00008 \quad \text{K}^{-1}\][/tex]

Now plug these values back into the Clausius-Clapeyron equation:

[tex]\[0.388 = \frac{\Delta H_{\text{vap}}}{8.314} \times 0.00008\][/tex]

Solve for [tex]\(\Delta H_{\text{vap}}\):[/tex]

[tex]\[\Delta H_{\text{vap}} = \frac{0.388 \times 8.314}{0.00008}\][/tex]

[tex]\[\Delta H_{\text{vap}} = \frac{3.226}{0.00008}\][/tex]

[tex]\[\Delta H_{\text{vap}} \approx 40325 \text{ J/mol} = 40.3 \text{ kJ/mol}\][/tex]

So, the heat of vaporization [tex](\(\Delta H_{\text{vap}}\))[/tex] of pinene is approximately 40.3 kJ/mol.

Answer:

The structure analysis says the compound must be Cumene or isopropylbenzene

Explanation:

Degree of unsaturation or double bond equivalent

         D.B.E     =      [tex]C-\frac{H}{2}+\frac{N}{2}+1[/tex]

                       =       [tex]9-\frac{12}{2}+\frac{0}{2}+1[/tex]

                       =       4

¹H NMR data analysis

  (i)  1.2δ (doublet, I = 6H)   two CH₃ are equivalents and the multiplicity says the neighboring carbon have one hydrogen.

 (ii)  3.0δ (septet, I = 1H), one CH and the multiplicity says the neighboring carbon  have six hydrogens.

 (iii)  7.1δ (multiplet, I = 5H) , means

and the sturcture of the compound is

Final answer:

Based on the 1H NMR spectroscopic data, the structure of the compound with the molecular formula C9H12 is determined to likely be ethylbenzene, featuring an aromatic ring with an attached ethyl group.

Explanation:

The question involves determining the structure of a compound with a molecular formula of C9H12 based on its 1H NMR spectroscopic data. The data provided are 1.2δ (doublet, I=6H), 3.0δ (septet, I=1H), and 7.1δ (multiplet, I=5H). To solve this, we analyze each piece of information. The doublet at 1.2δ with 6 hydrogens suggests the presence of two identical methyl groups (CH3) next to a carbon that splits their signal into a doublet. The septet at 3.0δ indicates a methine group (CH) that is adjacent to six hydrogens, likely from two methyl groups, causing this splitting pattern. Finally, the multiplet at 7.1δ for 5 hydrogens indicates the presence of an aromatic ring. Taking all this into account, a plausible structure for this compound is ethylbenzene, which consists of a benzene ring with an ethyl group attached to it.

LeFinal answer:

Based on the 1H NMR spectroscopic data, the structure of the compound with the molecular formula C9H12 is determined to likely be ethylbenzene, featuring an aromatic ring with an attached ethyl group.



Explanation:

The question involves determining the structure of a compound with a molecular formula of C9H12 based on its 1H NMR spectroscopic data. The data provided are 1.2δ (doublet, I=6H), 3.0δ (septet, I=1H), and 7.1δ (multiplet, I=5H). To solve this, we analyze each piece of information. The doublet at 1.2δ with 6 hydrogens suggests the presence of two identical methyl groups (CH3) next to a carbon that splits their signal into a doublet. The septet at 3.0δ indicates a methine group (CH) that is adjacent to six hydrogens, likely from two methyl groups, causing this splitting pattern. Finally, the multiplet at 7.1δ for 5 hydrogens indicates the presence of an aromatic ring. Taking all this into account, a plausible structure for this compound is ethylbenzene, which consists of a benzene ring with an ethyl group attached to it.

Answer:

[tex]T_F=31.9^0C[/tex]

Explanation:

Hello,

In this case, the energy content is modeled via:

[tex]Q=mCp(T_F-T_0)[/tex]

Since the unknown is the final temperature, it is solved by:

[tex]T_F=T_0+\frac{Q}{mCp} =33.9^0C+\frac{-69.0kJ}{8.10kg*4.18\frac{kJ}{kg*^0C}} \\T_F=31.9^0C[/tex]

The heat is negative since it flows out of the bath that is the considered system.

Best regards.

Answer:

The correct answers  "a) second order reaction." and "b) The first step is the slow step."

Explanation:

Order of NO = 1

Order of Cl₂ = 1

Overall order = 1 + 1 = 2

∴ the order is a second order reaction

Mechanism of the given reaction:

NO(g) + Cl₂(g) → NOCl₂(g)  ....................... Slow Step

NOCl₂(g) + NO(g) → 2NOCl(g)  ................ Fast step

The first step is the determining step, which is the slow step

The incorrect statements are:

"c) Doubling [NO] would decrease the rate by a factor of two. "

        When the concentrations of [NO] is doubled, the rate of the reaction is increased by the factor of two.

"d) The molecularity of the first step is 1. "

        Since the overall reaction is 2, the molecularity of the first step is 2.

"e) Both steps are termolecular."

         No, because the first step is bimolecular step.

Final answer:

The rate law for the reaction is R=k[NO][Cl2]. The first step is the slow step and the molecularity of the first step is 2. Doubling [NO] would increase the rate by a factor of four.

Explanation:

The rate law for the reaction 2NO(g) + Cl2(g) → 2NOCl(g) is given by R=k[NO][Cl2]. According to the given mechanism for the reaction, the first step is the slow step. Therefore, option b) is correct. The molecularity of the first step is 2 because it involves the collision of two molecules (NO and Cl2). Therefore, option e) is incorrect. Doubling [NO] would increase the rate by a factor of four (2^2), not decrease it by a factor of two. Therefore, option c) is incorrect. Hence, the correct options are b) and d).

To convert a temperature of 31 degrees Celsius to the equivalent temperature in degrees B on the undiscovered planet, we must use the given ratios of temperature differences, resulting in a converted temperature of 90.3 degrees B.

The temperature range for water between freezing and boiling on this undiscovered planet expressed in degrees B is 130 (180-50). The same range on Earth in degrees Celsius is 100 (100-0). First, we need to find the ratio of these two scales. The ratio is 130/100 = 1.3.

Next, to convert the Earth temperature from Celsius (31 °C) to degrees B, we multiply by the ratio and add the freezing point in degrees B. This gives us: (31 * 1.3) + 50 = 90.3 °B. So, 31 °C on Earth would be 90.3 °B on this undiscovered planet.

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Answer:

kudos to this man hope he is right if not sorry for taking the slot

Explanation:

Final answer:

Kp for the reactions at 873 K can be found by using the reciprocal, multiplication, or division based on the stoichiometry changes from the original reaction. For Part A, Kp is 3.45×10⁴; for Part B, it is 2.43×10⁻; and for Part C, it is 8.41×10⁻.

Explanation:

To find the value of Kp for the given reactions at 873 K, we use the reciprocal, multiplication, or division of the original reaction's Kp value depending on the stoichiometry of each reaction.

Part A

For the reaction 2AsH3(g)+P2(g)⇌2PH3(g)+As2(g), which is the reverse of the given reaction, the equilibrium constant Kp is the reciprocal of the original reaction's Kp. So, Kp for this reaction is 1 / (2.9×10⁻⁵) = 3.45×10⁴.

Part B

For the reaction 6PH3(g)+3As2(g)⇌3P2(g)+6AsH3(g), which is the original reaction multiplied by 3, the equilibrium constant Kp is the original Kp raised to the power of 3. Therefore, Kp is (2.9×10⁻⁵)³ = 2.43×10⁻.

Part C

For the reaction 2P2(g)+4AsH3(g)⇌2As2(g)+4PH3(g), which is the original reaction multiplied by 2, the equilibrium constant Kp is the original Kp squared. Thus, Kp for Part C is (2.9×10⁻⁵)² = 8.41×10⁻.

The given question is incomplete. The complete question is as follows.

The boiling points for a set of compounds in a homologous series can be qualitatively predicted using intermolecular force strengths. Using their condensed structural formulas, rank the homologous series for a set of alkanes by their boiling point.

Rank from highest to lowest boiling point. To rank items as equivalent, overlap them.

butane ([tex]C_{4}H_{10}[/tex]), 3,3-dimethylpentane ([tex]C_{7}H_{16}[/tex]), hexane ([tex]C_{6}H_{14}[/tex]), and heptane ([tex]C_{6}H_{16}[/tex]).

Explanation:

It is known that boiling point is the temperature at which vapor pressure of a liquid becomes equal to the atmospheric pressure. In hydrocarbons, more linearly the carbon atoms are attached to each other more will be the boiling point of the compound because of increase in surface area of the compound.

And, more is the branching present in a compound lesser will be its boiling point.

Also, more is the intermolecular strength present in a compound more will be its boiling point.

Thus, we can conclude that given compounds are ranked from highest to lowest boiling point as follows.

          heptane > hexane > 3,3-dimethyl pentane > butane

Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = [tex]25^{o} C[/tex]

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = [tex]C \times 0.082 \times 298.15 K [/tex]

               C = 0.0285

This also means that,

  [tex]\frac{\text{moles}}{\text{volume (in L)}}[/tex] = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = [tex]2.85 \times 10^{-3 }[/tex]

Now, let us assume that mass of [tex]C_{12}H_{23}O_{5}N[/tex] = x grams

And, mass of [tex]C_{12}H{22}O_{11}[/tex] = (1.00 - x)

So, moles of [tex]C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}[/tex]

                              = [tex]\frac{x}{369}[/tex]

Now, moles of [tex]C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}[/tex]

                   = [tex]\frac{x}{369} + \frac{(1.00 - x)}{342}[/tex]

                  = [tex]2.85 \times 10^{-3}[/tex]

             = x = 0.346

Therefore, we can conclude that amount of [tex]C_{12}H_{23}O_{5}N[/tex] present is 0.346 g  and amount of [tex]C_{12}H_{22}O_{11}[/tex] present is (1 - 0.346) g = 0.654 g.

Answer:

For the first reaction the reagents are: MgCl2 and Na2CO3

For the second reaction the reagents are: Na2HPO4 and CaCl2

Explanation:

Precipitation reactions lie in the production of a compound that is not soluble, which is called a precipitate, this precipitate is produced when two different solutions are combined, each of which will contribute an ion for the formation of the precipitate. In the first reaction you have:

MgCl2 + Na2CO3 = MgCO3 + 2 NaCl

Type of reaction: double displacement

The second reaction is as follows:

4Na2HPO4 + 3CaCl2 → Ca3 (PO4) 2 + 2NaH2PO4 + 6NaCl

It is the reaction of sodium hydrochlorophosphate and calcium chloride

Answer:

a.

C₂H₂ (g) + 5/2 O₂ (g) ⇒   2CO₂ (g) +   H₂O (l)

b. 6.21 g H₂O

c. 1.08 g

d. 6.21 g H₂O

e. 16 %

Explanation:

This question involves a calculation based on the stoichiometry of the balanced chemical equation:

b.

Lets calculate the # moles C₂H₂ 8.98 g will represent and then calculate the amount of water produced as follows:

# moles C₂H₂  = mass/molar mass = 8.98 g / 26.04 g/mol = 0.34 mol

From the stoichiometry of the reaction:

1 mol H₂O produced / mol C₂H₂  x 0.34 mol C₂H₂  = 0.34 mol H₂O produced

g H₂O = # mol H₂O x molar mass H₂O = 0.34 mol x 18.01 g/mol = 6.21 g H₂O

c.

For 4.58 g O₂ we can calculate the amount of water in grams formed as follows:

# mol O₂ = mass / molar mass O₂ = 4.58 g / 32 g / mol = 0.14 mol

From the stoichiometry of the reaction we have

1 mol H₂O produced /2.5 mol O₂ x 0.14 mol O₂ = 0.06 mol H₂O

mass H₂O produced = 0.06 mol x molar mas H₂O = 0.06 mol x 18.01 g/mol

                                  = 1.08 g H₂O

d,e.

We  calculated in part b  that we should have produced 6.21 g H₂O, therefore the percent yield is =

1 g / 6.21  g x 100 g = 16 %

Note one could argue that this theoretical yield refers to the 4.58 grams O₂ in part c. However if that were the case we will have more than 100 % yield, unless we round the numbers to give us 100 % yield

Final answer:

To find the grams of water formed from acetylene or oxygen, we use stoichiometry, converting grams to moles and using the balanced equation to relate substances. The theoretical yield is the maximum product amount if all the limiting reactant is used, and the percent yield is actual over theoretical yield times 100%.

Explanation:

The balanced chemical equation for the combustion of acetylene (C2H2) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:  2C2H2(g) + 5O2(g) ightarrow 4CO2(g) + 2H2O(l)

To calculate the gram of water formed from the consumption of acetylene or oxygen, we'll use stoichiometry. For instance, from 8.98 grams of acetylene, you first determine the molar mass of C2H2 (26.04 g/mol) and convert the mass of C2H2 to moles. Then, you use the balanced equation to find the mol-to-mol ratio between acetylene and water, leading you to the moles of water. Finally, convert moles of water to grams using the molar mass of water (18.015 g/mol).

For the case of 4.58 grams of oxygen, you would do a similar calculation. Determine the molar mass of O2 (32.00 g/mol), convert the mass of O2 to moles, and use the stoichiometric ratios from the balanced equation to find the moles and then grams of water produced.

The theoretical yield is the maximum amount of product that could be formed from the given amounts of reactants, assuming complete conversion and that all of the limiting reactant is consumed.

The percent yield is calculated as the actual yield divided by the theoretical yield, multiplied by 100%. Therefore, to find the percent yield given an actual yield of 1.00g of water, you would need to compare it to the theoretical yield previously calculated.