Potassium reacts with oxygen gas to produce potassium oxide. Determine the percent yield for the reaction between 8.92 grams of potassium and 3.28 grams of oxygen gas if 6.36 grams of potassium oxide are produced?

Answers

Answer 1

Answer:

Percent Yield K₂O  =  59.2%

Explanation:

Before you can calculate the percent yield, you need to find the theoretical yield. This value is the amount of product produced using the balanced chemical equation and molar masses. However, the question does not specify which reactant is the limiting reagent. Therefore, you have to calculate the mass of the product starting from both values.

To find the theoretical yield, you need to (1) convert grams K/O₂ to moles K/O₂ (via molar masses), then (2) convert moles K/O₂ to moles K₂O (via mole-to-mole ratio from balanced equation coefficients), and then (3) convert moles K₂O to grams K₂O (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given values.

Molar Mass (K): 39.098 g/mol

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

Molar Mass (K₂O): 2(39.098 g/mol) + 15.998 g/mol

Molar Mass (K₂O): 94.194 g/mol

The balanced equation:

4 K + O₂ -----> 2 K₂O

8.92 g K          1 mole             2 moles K₂O         94.194 g
--------------  x  -----------------  x  --------------------  x  ----------------  =  10.7 g K₂O
                       39.098 g           4 moles K             1 mole

3.28 g O₂          1 mole           2 moles K₂O         94.194 g
----------------  x  ---------------  x  ---------------------  x  --------------  =  19.3 g K₂O
                         31.996 g           1 mole O₂              1 mole

Because potassium produces the smaller amount of product, it is the limiting reagent. This means that all the potassium reactant is used up before the oxygen gas runs out. So, the theoretical yield of potassium oxide is 10.7 grams.

Now, you can use the theoretical yield and actual yield to determine the percent yield.


                                   Actual Yield
Percent Yield  =  ------------------------------  x  100%
                               Theoretical Yield

                                 6.36 g K₂O
Percent Yield  =  ------------------------  x  100%
                                 10.7 g K₂O

Percent Yield  =  59.2%


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How many grams of silver may be formed by the passage of 9,604 c through an electrolytic cell that contains a molten silver salt?

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Answers

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Answer:

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2.18 mg of naoh is needed to prepare 546 ml of solution with a ph of 10.00.

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Answer:

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It is refrigeration's coefficient of performance (COP) will always be greater than 1.

Therefore, the coefficient of performance (cop) of a refrigerator is defined as the ratio of " the work necessary to heat or cool something usefully."

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A solution of naf is added dropwise to a solution that is 0.0122 m in ba2 . when the concentration of f- exceeds ________ m, will precipitate. neglect volume changes. for , ksp = 1.7 x 10-6. 4. calculate the ph of a solution prepared by dissolving 0.270 mol of weak acid ha (ka = 1.77 x 10-4) and 0.260 mol of its conjugate base in water sufficient to yield 1.00 l of solution.

Answers

A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 m, BaF₂ will precipitate.

Let's consider the solution of BaF₂.

BaF₂(s) ⇄ Ba²⁺(aq) + 2 F⁻(aq)

We can use the solubility product constant (Ksp) to find the equilibrium concentration of F⁻ when [Ba²⁺] is 0.0144 M.

When [F⁻] exceeds 0.011 M, BaF₂ will precipitate.

A solution of NaF is added to a solution that is 0.0144 M in Ba²⁺. When the concentration of F- exceeds 0.011 M, BaF₂ will precipitate.

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How much 0. 5mna2so4 solution will completely precipitate the ba2 in 0. 7l of 0. 13mbacl2 solution?

Answers

The volume required to precipitate 0.7l of 0.13 m BaCl2 is 0.04L.

Calculations of number of moles for given molarity of solution

we use

Molarity =Moles of solute / volume of solution (in litre)

Now, for barium ions

Given,

Molarity of solution = 0.13 m

Volume of solution = 0.7L

By substituting all the value ,we have

Moles of Barium = 0.7 × 0.13

= 0.091

Chemical equation

BaSO4 » Ba2+ + So2-

By stoichiometry of equation

1 mole of sulphate ion precipitate 1 mole of barium ion

So, 0.091 mole of sulphate ion precipitate

0.091 of moles of Barium ion

Calculation of volume of sulfate ion

Now, calculate the volume of ions by using equation as given above

Mole of sulphate ion = 0.091 mol

Molarity concentration of sulphate ions = 0.05m

By substituting all value , we get

C = n / V

V = n/ C

= 0.091 / 0.5 = 0.04L

Thus, we calculated that the volume required to precipitate 0.7l of 0.13 m BaCl2 is 0.04L.

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Bohr's model explained the position of the electron, proton, and neutron in the atom of the element. The energy at the n = 2 level of the atom will be - 3.40 eV.

What is the principal quantum number (n)?

The principal quantum number (n) has been the distance of the electron of that atom in the nucleus and its energy in the structure. It can also be said to define the size of the atomic orbit.

n = 2 is the first excited state whose energy is calculated as:

Eₙ = − 13.6 ÷ n² eV

E₂ = - 13.6 eV ÷ 2²

= -3.40 eV

Therefore, -3.40 eV is the energy of electron at n = 2.

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Simon measured the density of a piece of metal
to be 11.70 g/cm³. However, the manufacturer
claims it has a density of 12.00 g/cm³. What is
the percent error of Simon's measurement?

Answers

From the question we know that 12 - 11.70 = 0.3  so 0.3 / 12 = 0.025 therefore Simon's percent error is 2.5%.

Percent error compares an estimate to an accurate value and expresses the difference among them as a percent. This statistic lets analysts recognize the scale of the mistake relative to the true cost. it's also referred to as percentage error as well as % mistakes.

Percentage errors Calculation Steps :

Subtract one cost from another.Divide the mistake by the exact or ideal price (not your experimental or measured value).Convert the decimal variety into a percentage by multiplying it by using 100.Add a percent or % symbol to record your percentage mistakes value.

Percentage error tells you the way huge your mistakes are when you degree something in an experiment. Smaller values suggest that you are close to the accepted or actual value. for instance, a 1% blunder means that you obtain very near the accepted value, whilst 45% means that you were pretty a long way off from the real value.

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3h2so4 2b(oh)3 - b2(so4)3 6h2o what is the mole ratio between water and sulfuric acid?

Answers

Since there is an equal number of each element in the reactants and products of 3H2SO4 + 2B(OH)3 = B2(SO4)3 + 6H2O, the equation is balanced.

What is Sulfuric acid?Sulfuric acid (H2S04) is a corrosive substance, that destroys the skin, eyes, teeth, and lungs. Severe vulnerability can result in death. Workers may be harmed from disclosure to sulfuric acid. The level of exposure depends on The sulfuric acid bleaching can substitute the presently assumed oxygen and chlorine stages if the additives are allowed. This bleaching process was also practical for oxygen-bleached hardwood kraft pulp, but it was less compelling for softwood kraft pulp and oxygen-bleached softwood kraft pulp.Disbanding sulfuric acid in water is an exothermic procedure. When sulfuric acid is mixed with ice this exothermic procedure causes the temperature to rise at first, but as more of the ice dissolves, the temperature falls.

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In the laboratory, a student dilutes 18. 9 ml of a 10. 0 m perchloric acid solution to a total volume of 250. 0 ml. what is the concentration of the diluted solution? concentration = m submit answer

Answers

The concentration of diluted solution is 0.756M.

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 18.9 mL

Molarity of stock solution (M1) = 10 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (M2) =?

We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:

M1V1 = M2V2

10 × 18.9 = M2 ×250

189 = M2 × 250

Divide both side by 100

M2 = 189 / 250

M2 = 0.756 M

Therefore, the molarity of the diluted solution is 0.756 M.

Thus the concluded that concentration of the dilute acid is 0.756 M.

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What substance is produced at the cathode during the electrolysis of molten calcium bromide, cabr2cabr2? assume standard conditions

Answers

Magnesium (Mg) metal is the substance produced at the cathode.

As molten calcium bromide is electrolyzed, calcium (Ca) metal is the substance produced at the cathode.

Due to the fact that oxidation always takes place at the anode and reduction always occurs at the cathode,

Consider the two cations that were produced in the two ionisation processes of the two substances: mathrm Ca +2 and mathrm Mg +2. The following is how to express two cations' potential cathode reactions:

[tex]ca { }^{ + 2} + (aq) + 2e {}^{ - } - > \\ ca(s)E {}^{0} _{ca {}^{ + 2} \div ca } = - 2.87v[/tex]

[tex] Mg{ }^{ + 2} + (aq) + 2e {}^{ - } - > \\Mg(s)E {}^{0} _{Mg {}^{ + 2} \div Mg} = - 2.37v[/tex]

[tex] Mg^{ + 2} [/tex]

will be reduced quickly since it has a low electronegative electrode potential.

Thus, magnesium (Mg) metal is the substance produced at the cathode.

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Hello people ~
Volume of one mole of any gas at NTP is
(a) 11.2 litre
(b) 22.4 litre
(c) 10.2 litre
(d) 22.8 litre

Answers

Answer:

Volume of one mole of any gas at NTP is

(b) 22.4 litre

Answer:

22.4L

Explanation:

One mole of any gas at STP weighs 22.4L.

STP stands for Standard Temperature and pressure

273.15KPressure 1atm

What is the molar mass of a protein if a 0. 30 l of solution containing 0. 45 g of the protein has an osmotic pressure of 0. 80 torr at 25∘c?

Answers

The molar mass of a protein if a 0. 30 l of solution containing 0. 45 g of the protein has an osmotic pressure of 0. 80 torr at 25∘c is 0.348 g/mol

Here we can apply the formula ∏ = iMRT, where ∏ = osmotic pressure = 0.80 - ( given ). This is only one part of the information we are given / can conclude in this case ....

i = van’t Hoff factor = 1 for a protein molecule,

R = gas constant = 62.36 L torr / K-mol,

T ( temperature in Kelvin ) = 25 + 273 - conversion factor C° + 273 = 298K

( Known initially ) ∏ = osmotic pressure = 0.80 torr

besides the part " M " in the formula, which we have no information on whatsoever, as we have to determine it's value.

_____

Substitute derived / known values to solve for M ( moles / liter ) -

∏ = iMRT

⇒ 0.80 = ( 1 )( M )( 62.36 )( 298 )

⇒ 0.80 = M( 18583.28 )

⇒ M = 0.80 / 18583.28 ≈ 4.3049 ....

_____

We know that M = moles / liter, so we can use this to solve for moles, and hence calculate the molar mass by the formula molar mass = g / mol -

M = mol / l

⇒ 4.3049 = 0.045 / 25 mL ( 0.030 L ),

0.045 / 0.030 = 1.5 g / L

⇒ 1.5 g = 4.3049 moles,

molar mass = 1.5 g = 4.3049 moles = 0.348 g / mol

Thus the concluded that the molar mass of protein is 0.348 g / mol.

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How many particles are there in 5 grams of sodium carbonate?

Answers

3.01× 1024 particles are the number of  particles are there in 5 grams of sodium carbonate.

How many particles are there in 5 grams of sodium carbonate?

There are 6.022 × 1023 particles in one gram of a substance according to Avogadro's number. So when we find out for 5 grams, then we multiply 5 with 6.022 × 1023, we get 3.01 × 1024 particles. For one gram atomic weight of hydrogen, one mole of hydrogen contains 6.022 × 1023 hydrogen atoms.

So we can conclude that 3.01× 1024 particles are the number of  particles are there in 5 grams of sodium carbonate.

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