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erastova [34]
3 years ago
7

What is the most important reason to publish results as part of the scientific process?

Chemistry
2 answers:
Makovka662 [10]3 years ago
8 0

It allows other scientists to check findings.

svp [43]3 years ago
4 0
<span>It allows other scientists to check findings</span> is the most important reason to publish results as part of the scientific process.
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Some help I’ll give brainliest
daser333 [38]

Answer: Meteoroid

Explanation:

Dwarf planets are smaller than moons

Comets are smaller than dwarf planets.

Meteoroids, Meteors, and Meteorites are broken off pieces of comets.

Meteoroids are smaller than Asteroids

5 0
3 years ago
Which ion in the ground state has the same electron configuration as an atom of neon in the ground state?
Ainat [17]
  • Cl-

Lets see how

  • Z for Cl is 9
  • One electron is added so new Z=10

Electronic configuration

  • 1s²2s²2p⁶

Or

  • [Ne]

option B is correct

6 0
2 years ago
Read 2 more answers
Which type of wave do the particles in a medium move parallel to the direction that the waves move
Talja [164]
Longitudinal waves. In a longitudinal wave the particles in the medium move parallel  to the direction waves go. A good example can be the p-waves in an earthquake.
8 0
4 years ago
Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to
vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

   H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)

Next we will calculate how many moles of H_2SO_4 are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = \frac{\text{moles of solute}}{\text{liters of solution&#10;}}

            1.500 M = \frac{n}{0.08500 L&#10;}

                    n = 0.1275 mol H_2SO_4

Now set up and solve a stoichiometric conversion from moles of H_2SO_4  to grams of NaHCO_3. As, the molar mass of NaHCO_3 is 84.01 g/mol.

 0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})

                 = 21.42 g NaHCO_3

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0
3 years ago
A nail contains 0.10 moles of iron. How many grams of iron are in the nail?
oksano4ka [1.4K]

Answer:

10.8

Explanation:

That's my guess, I hope you figure everything out though.

4 0
3 years ago
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