CHEMISTRY
MIDDLE SCHOOL

Answer:

Scientists are likely to accept a new or modified theory if it explains everything the old theory did and more. The process of theory change may take time and involve controversy, but eventually the scientific explanation that is more accurate will be accepted.

Hope this helps

HIGH SCHOOL

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook). Part A. 0.270 M

Part B. 7.84×10-2 M

Part C. 1.97×10-2 M

The percent ionization for propionic acid in each case are:

- Part A. 0.704%
- Part B. 1.31%
- Part C. 2.61 %

Percent ionization is the quantity of a weak acid that ionizes in a solution expressed as a percentage.

We can calculate the percent ionization using the following expression.

α% = √(Ka/Ca) × 100%

where,

- α% is the percent ionization.
- Ka is the acid dissociation constant (1.34 × 10⁻⁵ for propionic acid).
- Ca is the concentration of the acid.

- Part A. 0.270 M

α% = √(1.34 × 10⁻⁵/0.270) × 100% = 0.704%

- Part B. 7.84 × 10⁻² M

α% = √(1.34 × 10⁻⁵/7.84 × 10⁻²) × 100% = 1.31 %

- Part C. 1.97 × 10⁻² M

α% = √(1.34 × 10⁻⁵/1.97 × 10⁻²) × 100% = 2.61 %

The percent ionization for propionic acid in each case are:

- Part A. 0.704%
- Part B. 1.31%
- Part C. 2.61 %

Learn more about percent ionization here: brainly.com/question/20629694

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

=

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)

percentage% (∝) =

= 0.704%

For Part B; where Concentration of B = M

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C=

percentage% (∝) = 0.02608 × 100%

= 2.60%

MIDDLE SCHOOL

Name one common misconception of thinking the Earth is closet to the sun during the summer

Earth is closest to the Sun when it's summertime or, it is warmer while in the summer

COLLEGE

Write the balanced chemical equation for the complete, stoichiometric combustion of ethylene in (a) nitrous oxide and (b) air. Compare the required number of moles and the oxidizer mass using each of the two oxidizers for the complete, stoichiometric combustion of ethylene.

Answer: Mass of nitrous oxide used is 264 grams and mass of oxygen gas used is 96 grams.

Explanation:

To calculate the number of moles, we use the equation:

....(1)

- For a: Combustion of ethylene in nitrous oxide

The chemical equation for the combustion of ethylene in nitrous oxide follows the equation:

By stoichiometry of the reaction;

6 moles of nitrous oxide are required for the complete combustion of ethylene molecule.

Calculating the mass of nitrous oxide using equation 1:

Molar mass of nitrous oxide = 44 g/mol

Moles of nitrous oxide = 6 moles

Putting values in equation 1, we get:

Thus, 264 grams of nitrous oxide are required for the complete combustion of ethylene.

- For b: Combustion of ethylene in air

The chemical equation for the combustion of ethylene in air follows the equation:

By stoichiometry of the reaction;

3 moles of oxygen are required for the complete combustion of ethylene molecule.

Calculating the mass of oxygen using equation 1:

Molar mass of oxygen = 32 g/mol

Moles of oxygen = 3 moles

Putting values in equation 1, we get:

Thus, 96 grams of oxygen are required for the complete combustion of ethylene.

HIGH SCHOOL

What is the difference between "4 P" and "P4"? A) There is no difference, they both represent the same thing. B) 4 P represents four molecules and P4 represents four atoms. C) P4 represents four atoms chemically bonded together and 4 P does not. D) 4 P represents phosphorus atoms that are bonded together chemically and P4 represents one molecule

The answer, I think, would be "C". This is because 4P represents that there are 4 separate atoms of phosphorous; opposingly, P4 indicates that there are 4 atoms in a single molecule of phosphorous (that have chemically bonded together). Hope this helps!

HIGH SCHOOL

What is the volume of 12.8 g of a liquid that has a density of 0.789 g/ml? 13.6 ml 12.8 ml 16.2 ml 10.7 ml none of the above?

Answer: The volume of liquid is 16.2 mL

Explanation:

To calculate volume of a substance, we use the equation:

We are given:

Density of liquid = 0.789 g/mL

Mass of liquid = 12.8 g

Putting values in above equation, we get:

Hence, the volume of liquid is 16.2 mL

Density=mass/volume volume=mass/density volume=12.8g/0.789 volume=16.2ml

HIGH SCHOOL

What type of crystal will each of the following substances form in its solid state? (metallic, ionic, covalent)

I think these are the substances that are needed to be categorized into ionic, metallic or covalent: C2H6, Na2O, SiO2, CO2, N2O5, NaNO3, Al, C (diamond) and SO2. Ionic substances are Na2O, SiO2, NaNO3; covalent substances are C2H6, CO2, N2O5, C (diamond); and metallic substances are Al, SO2.

COLLEGE

The decomposition reaction of carbon disulfide to carbon monosulfide and sulfur is first order with k = 2.80 ✕ ✕ 10−7 sec-1 at 1000°c. cs2(g) → cs(g) + s(g) a. how much of a 4.83-gram sample of carbon disulfide would remain after 37.0 days? 1.97 1.97 grams carbon disulfide

b. how much carbon monosulfide would be formed after 37.0 days? 1.14 1.65 grams carbon monosulfide useful information 1.013 bar = 760 torr = 1 atm = 760 mm hg

The first order reaction for decomposition of carbon disulfide is as follows:

The rate constant for the above reaction is .

(a) The expression for rate constant for first order reaction is as follows:

Here, k is rate constant, t is time of the reaction, is initial concentration of reactant and is concentration at time t.

Here, concentration terms can be replaced by mass now, the mass of reactant at time 37 days should be calculated.

First convert unit of time from days to sec as follows:

1 day=86400 sec

thus,

Now, putting the values in expression for rate constant,

On rearranging,

Taking antilog both sides,

Or,

Therefore, mass of carbon disulfide remains after 37 days is 1.97 g.

(b) From the balanced chemical equation for decomposition of carbon disulfide 1 mol of gives 1 mol of CS.

now, mass of carbon disulfide remain after 37 days is 1.97 g thus, mass of carbon disulfide reacted can be calculated as follows:

Calculating number of moles from it,

Molar mass of is 76.14 g/mol

Putting the values,

Since, 1 mol of gives 1 mol of CS thus, 0.0375 mol will give 0.0375 mol of CS.

Molar mas of CS is 44.07 g/mol , calculate mass as follows:

Therefore, mass of carbon monosulfide formed after 37 days is 1.65 g.

MIDDLE SCHOOL

What could go wrong in a science lab?

Someone could spill something

HIGH SCHOOL

How many molecules of ethanol (c2h5oh) (the alcohol in alcoholic beverages) are present in 140 ml of ethanol? the density of ethanol is 0.789 g/cm3?

Taking into account the definition of density and Avogadro's number, 1.44×10²⁴ molecules of ethanol are present in 140 ml of ethanol.

Definition of densityDensity is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

Avogadro's Number

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

Amount of molecules of ethanolIn this case, you know that:

- Density= 0.789
- Volume= 140 mL= 140 cm³

Replacing in the definition of density:

Solving:

mass= 0.789 ×140 cm³

mass= 110.46 g

The molar mass of ethanol, that is, the amount of mass present in one mole of the compound, is 46 . Then the number of moles that 110.46 g of ethanol contain is calculated by:

110.46 g×= 2.40 moles

Finally, you can apply the following rule of three: If by definition of Avogadro's number 1 mole of ethanol contains 6.023×10²³ molecules, 2.40 moles contains how many molecules?

amount of molecules= (2.40 moles× 6.023×10²³ molecules)÷ 1 mole

amount of molecules= 1.44×10²⁴ moles

In summary, 1.44×10²⁴ molecules of ethanol are present in 140 ml of ethanol.

Learn more about

density:

brainly.com/question/952755?referrer=searchResults

brainly.com/question/1462554?referrer=searchResults

Avogadro's Number:

brainly.com/question/1445383?referrer=searchResults

brainly.com/question/1528951?referrer=searchResults

An mL is also equivalent to cm³, signifying that 140 mL is equivalent to 140 cm³. The mass of ethanol is calculated by multiplying the density by the volume.

mass = (140 cm³)(0.789 g/cm³)

mass = 110.46 g

Then, calculate the number of moles of ethanol by dividing the mass by the molar mass of ethanol equal to 46.07 g/mol.

number of moles = (110.46 g/ 46.07 g/mol) = 2.4 mol

Then, multiply the number of moles by the Avogadro's number.

2.4 mol (6.022 x 10²³)

number of molecules = 1.445 x 10²⁴

mass = (140 cm³)(0.789 g/cm³)

mass = 110.46 g

Then, calculate the number of moles of ethanol by dividing the mass by the molar mass of ethanol equal to 46.07 g/mol.

number of moles = (110.46 g/ 46.07 g/mol) = 2.4 mol

Then, multiply the number of moles by the Avogadro's number.

2.4 mol (6.022 x 10²³)

number of molecules = 1.445 x 10²⁴