Chemistry, 14.03.2020 04:25 heavenwagner

3. What was your average percent error for all three trials? In this experiment you did three trials. Why was it suggested that you do three trials and not fewer?

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Answer from: Aethis

To be sure of the value you will obtain

Explanation:

Trial 1

Mass of calorimeter: 409g

Temp of calorimeter: 20Cº

Mass of calorimeter + water: 457g

Temp of calorimeter + water: 20Cº

Mass of calorimeter + water + ice: 510g

Temp of calorimeter + water + ice: –

Mass of calorimeter + ice: 450g

Temp of calorimeter + ice: 0Cº

Trial 2

Mass of calorimeter: 409g

Temp of calorimeter: 20Cº

Mass of calorimeter + water: 461g

Temp of calorimeter + water: 22Cº

Mass of calorimeter + water + ice: 515g

Temp of calorimeter + water + ice: –

Mass of calorimeter + ice: 447g

Temp of calorimeter + ice: 0Cº

Trial 3

Mass of calorimeter: 409g

Temp of calorimeter: 20Cº

Mass of calorimeter + water: 459g

Temp of calorimeter + water: 22Cº

Mass of calorimeter + water + ice: 513g

Temp of calorimeter + water + ice: –

Mass of calorimeter + ice: 448g

Temp of calorimeter + ice: 0Cº

Answer from: isaiah1027

To be sure of the value you will obtain

Explanation:

Trial 1Mass of calorimeter: 409gTemp of calorimeter: 20CºMass of calorimeter + water: 457gTemp of calorimeter + water: 20CºMass of calorimeter + water + ice: 510gTemp of calorimeter + water + ice: –Mass of calorimeter + ice: 450gTemp of calorimeter + ice: 0Cº

Trial 2Mass of calorimeter: 409gTemp of calorimeter: 20CºMass of calorimeter + water: 461gTemp of calorimeter + water: 22CºMass of calorimeter + water + ice: 515gTemp of calorimeter + water + ice: –Mass of calorimeter + ice: 447gTemp of calorimeter + ice: 0Cº

Trial 3Mass of calorimeter: 409gTemp of calorimeter: 20CºMass of calorimeter + water: 459gTemp of calorimeter + water: 22CºMass of calorimeter + water + ice: 513gTemp of calorimeter + water + ice: –Mass of calorimeter + ice: 448gTemp of calorimeter + ice: 0Cº

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

can you list the answers so i can answer

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3. What was your average percent error for all three trials? In this experiment you did three trials...

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