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3 years ago

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

Chemistry

2 answers:

olganol [36]3 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]3 years ago

3 0

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

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

Which element, represented by X, reacts with oxygen to produce the compound X2O?

natali 33 [55]

Answer:

lithium

Explanation:

this is because lithium has a valency of 1 and oxygen has a valency of 2 thereby exchanging valency to create Li²0

7 0

3 years ago

Define the term relative atomic mass, A.

Naily [24]

The mass of an atom and the equivalent of to the number of protons and neutrons in the atom

6 0

2 years ago

what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

Law Incorporation [45]

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L