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4 years ago

Which is one way that scientist communicate the results of an experiment

Chemistry

2 answers:

olganol [36]4 years ago

4 0

Answer:

The way in which the scientist communicates the results of an experiment is through the publication of the scientific article in a scientific journal.

Explanation:

The way in which the scientist communicates the results of an experiment is through the publication of the scientific article in a scientific journal.

The publication of a scientific or technical article is a way of transmitting to the technical-scientific community the knowledge of new discoveries, and the development of new materials, techniques and methods of analysis in the various areas of science.

Many educational and / or research institutions, and several commercial companies often require their professionals to present the progress of their work and / or study through the publication of technical and scientific articles. In addition, institutions or companies that publish constantly enjoy the technical recognition of their name, which helps to attract greater investments and gains for this organization.

Slav-nsk [51]4 years ago

3 0

Draw conclusions,............

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You are using a standardized 0.1250 M solution of sodium hydroxide to determine the concentration of sulfuric acid. After pipett

Sergeu [11.5K]

Answer: Thus the concentration of the sulfuric acid is 0.1201 M

Explanation:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=10.00mL\\n_2=1\\M_2=0.1250M\\V_2=19.22mL$

Putting values in above equation, we get:

$2\times M_1\times 10.00=1\times 0.1250\times 19.22\\\\M_1=0.1201M$

Thus the concentration of the sulfuric acid is 0.1201

4 0

3 years ago

Which values do scientists compare when using potassium argon dating

bixtya [17]

The amount of parent and daughter material trapped in the rock.

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4 years ago

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GIVEN BRAINLIEST

Alla [95]

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3 years ago

When separating benzoic acid, naphthalene, and 3-nitroaniline, identify the solution used for each described purpose.

mamaluj [8]

Answer:

a. Extract reconstituted 3-nitroaniline out of the aqueous layer: Dichloromethane.

b. Extract benzoic acid into the aqueous layer: 5% NaOH.

c. Extract aniline into the aqueous layer: 5% HCl.

d. Reconstitute aniline from aqueous layer: 6M NaOH.

Explanation:

Each of the answers is explained as follows:

a. Extract reconstituted 3-nitroaniline out of the aqueous layer: Dichloromethane.

Dichloromethane is an organochloride chemical and CH2Cl2 is its formula. It is commonly employed as a solvent since it is a colorless, volatile liquid with a sweet, chloroform-like odor. It is polar and miscible with numerous organic solvents, despite not being miscible with water.

b. Extract benzoic acid into the aqueous layer: 5% NaOH.

According to the Mass by Volume percentage concept, 5% NaOH solution (W/V) indicates 5g of NaOH dissolved in 100ml of water.

c. Extract aniline into the aqueous layer: 5% HCl.

Hydrochloric acid (HCl) is a common acid found in the human body as well as in laboratories, and it is a key component of stomach acid.

As a result, 5% HCl solution is made by mixing 8.26 mL concentrated HCl with roughly 50 mL distilled water, stirring well, and then adding water up to 100 mL.

d. Reconstitute aniline from aqueous layer: 6M NaOH.

6M NaOH refer to 6 moles in 1L MW (NaOH). Therefore, 6 moles of NaOH are dissolved in 1 liter of H2O to make a 6M NaOH solution. Because NaOH has a molar mass of 39.9997 g/mol, we'll need 240 g NaOH (39.997 x 6) in 1L of H2O to make a 6M solution.

6 0

3 years ago

What is the molar mass of an unknown hydrocarbon whose density is measured to be 1.97 g/l at stp?

inessss [21]

Answer: The correct answer is : 44.1 g / mol

Explanation: d = PM / RT

M = dRT / P

d = 1.97 g / l

R = ideal gas (0.082 06 l - atm / mol - ° K)

P = 1 atm

T = 273 ° K

M = Molar mass

M = 1.97 g / l x 0.082 l - atm / mol - ° K x 273 ° K / 1 atm

M = 44.1 g / mol

5 0

3 years ago

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