What is the empirical formula of a compound composed of 36.7 g 36.7 g of potassium ( K K ) and 7.51 g 7.51 g of oxygen ( O O )? Insert subscripts as needed.

Answers

Answer 1
Answer:

Answer:

K₂O

Explanation:

Given parameters:

Mass of K = 36.7g

Mass of O = 7.51g

Unknown:

Empirical formula of the compound

Solution:

The empirical formula of a compound is it's simplest ratio by which the elements in the compound combines. It differs from the molecular formula that shows the actual atomic ratios.

To find the empirical formula, follow this process;

   

Elements                     K                               O

Mass                          36.7                            7.51

Molar

mass                            39                            16

Number of

moles                       36.7/39                    7.51/16

                                   0.94                        0.47

Divide by

the smallest              0.94/0.47               0.47/0.47

                                         2                              1

  Empirical formula is        K₂O

                                     

Answer 2
Answer:

Final answer:

The empirical formula of the compound composed of 36.7 g of potassium and 7.51 g of oxygen is K2O.

Explanation:

To determine the empirical formula of a compound, we need to find the ratio of the elements present. In this case, we have 36.7 g of potassium and 7.51 g of oxygen. To find the ratio, we need to convert these masses to moles by dividing them by the molar masses of potassium and oxygen. The molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Dividing the masses by the molar masses gives us 0.939 mol potassium and 0.469 mol oxygen. The ratio between these two elements is approximately 2:1, so the empirical formula of the compound is K2O.

Learn more about Empirical formula here:

brainly.com/question/32125056

#SPJ3


Related Questions

The following figure represents the formation of an ionic compound. Substances A and B are initially uncharged, but when mixed electrons are transferred. Using the figure, identify which substance will form the cation and which will form the anion. Provide a brief (one or two sentences) explanation for your response. (Hint: How does losing electrons affect atomic radii?)
3.What do we call materialsthat let heat pass throughthem easily?Thermal conductorsThermal insulatorsTransparent4.Which of these is a goodthermal conductor?PlasticWoodSteel5.Which of these is a goodthermal insulator?SteelIronPolystyrene6.To save on heating bills, doyou think the roof of abuilding should be lined with...a thermal conductora thermal insulatornothing7.How does heat travel?From cold things to hotter thingsFrom hot things to colder thingsBetween things of the same temperature
Draw all the structural isomers for the molecular formula C3H8O. Be careful not to draw any structures by crossing one line over another; the system needs to know where you intend connections to be between atoms.
(01.04 LC)What phase of matter has particles that are held together but can flow past eachother and takes the shape of a container, filling it from the bottom up? (3 points)1) Gas2) Liquid3) Plasma4) Solid
Name the process that happens when a liquid turns into a gas.​

Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is 2.5 x 10-14. Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is a.2.5 x 10-14 M.
b. 8.5 x 10-6 M
c. 6.3 x 10-11 M
d. 1.3 x 10-12 M
e. 5.0 x 10-2 M
f. 1.8 x 10-5 M

Answers

Answer:

c. 6,3x10⁻¹¹M

Explanation:

The solubility of a buffer is defined as the concentration of the dissolved solid in a saturated solution. For the Cd(OH)₂, solubility is:

[Cd²⁺] = S

The dissolution of Cd(OH)₂ is:

Cd(OH)₂ ⇄ Cd²⁺ + 2OH⁻

And the ksp is defined as:

ksp = [Cd²⁺][OH⁻]²

As ksp = 2,5x10⁻¹⁴ and [OH⁻] at pH=12,30 = 10^-(14-12,30) = 0,01995M

2,5x10⁻¹⁴ = [Cd²⁺]×(0,01995M)²

[Cd²⁺] = 6,3x10⁻¹¹M

That means solubility is c. 6,3x10⁻¹¹M

I hope it helps!

Final answer:

The molar solubility of Cd(OH)2 when buffered at a pH of 12.30 can be calculated using the concept of hydrolysis. The correct answer is 6.3 x 10^(-11) M.

Explanation:

To calculate the molar solubility of Cd(OH)2 when buffered at a pH of 12.30, we need to use the concept of hydrolysis. Cd(OH)2 is a slightly soluble salt that undergoes hydrolysis in aqueous solution. At a high pH value, OH- ions react with water to form more OH- ions, shifting the equilibrium towards the hydrolysis reaction.

  1. First, we write the balanced equation for the hydrolysis reaction: Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH-(aq)
  2. Since OH- is being produced, we can assume that the concentration of OH- is much greater than that of Cd2+. Therefore, we can ignore the concentration of Cd2+ when calculating the solubility product (Ksp).
  3. Next, we use the equation for the hydrolysis reaction to write the expression for the solubility product constant (Ksp): Ksp = [Cd2+][OH-]^2
  4. The concentration of OH- ions in a basic solution is related to the pH by the equation: pOH = 14 - pH
  5. Using this equation, we can calculate the pOH of the buffered solution: pOH = 14 - 12.30 = 1.70
  6. Then, we convert the pOH back to OH- concentration: [OH-] = 10^(-pOH) = 10^(-1.70)
  7. Finally, we substitute the calculated [OH-] into the expression for Ksp to solve for the molar solubility of Cd(OH)2: [Cd(OH)2] = sqrt(Ksp / [OH-]^2)

After performing the calculations, the molar solubility of Cd(OH)2 when buffered at a pH of 12.30 is approximately 6.3 x 10^(-11) M. Therefore, the correct answer is option c. 6.3 x 10^(-11) M.

Learn more about Molar solubility of Cd(OH)2 here:

brainly.com/question/30905557

#SPJ11

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

Answers

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?What is the PH of the solution?

Answers

Answer:

[H₃O⁺] = 2.5 × 10⁻¹³ M

pH = 12.6

Explanation:

Step 1: Given data

Concentration of OH⁻: 0.04 M

Step 2: Calculate the concentration of H₃O⁺

Let's consider the self-ionization of water reaction.

2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)

The ionic product of water is:

Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴

[H₃O⁺] = 10⁻¹⁴ / [OH⁻]

[H₃O⁺] = 10⁻¹⁴ / 0.04

[H₃O⁺] = 2.5 × 10⁻¹³ M

Step 3: Calculate the pH

The pH is:

pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6

lucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of approximately 0.200 mM. i) What is the concentration of glucose in the E. coli cell in mg/mL?

Answers

Answer:

The concentration is 0.036 mg/mL

Explanation:

Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL

Pls answer this question

Answers

Answer:

A carbon dioxide and oxygen

Convert a distance of 150 ft to its equivalent in cm

Answers

Answer:

4572cm

Explanation: 1ft =30.48cm so you multiply the length value by 30.48