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Wewaii [24]
3 years ago
6

1. Carson completed conductivity testing on the solutions shown in the data table below and recorded his findings in the conduct

ivity column. After determining which solutions conducted electricity and which solutions did not conduct electricity, Carson needs to explain his findings so that he can make predictions about conductivity in the future without having to complete the testing. Why do some of the solutions conduct electricity and some of the solutions not conduct electricity?
Results of conductivity testing: image

Consider Carson's findings, then make a claim to answer the question. Be sure to support your claim with sufficient evidence and reasoning.

Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

See below :)

Explanation:

There is an evident reason why some of the solutions Carson's has listed and observed, does conduct electricity and some that do.

A flow of electrical charge is called an electric current. Ions are atoms, or sets of atoms, that contain an electrical charge. There are two types of ions, cation or a positively charged ion containing a deficiency of electrons, and anion or a negatively charged ion which contains a surplus of electrons. When a solution conducts electricity the charge is carried within by ions that move through the solution. The larger the number of ions in the solution, the better the conductivity of the solution is. Pure water does not conduct very well because it contains very few ions, but when table salt (NaCl) is dissolved in the water, this solution does conduct well because the solution contains a more abundance of ions. The majority of the ions come from the table salt, chemically names sodium chloride. Because Sodium contains its sodium ions, and these are positive charge and chloride ions which is a negative charge, it is called an ionic substance. Not every substance is made up of ions, one such example is sugar (C12H22O11). Sugar is made up of uncharged particles also called molecules. Although sugar is a substance its molecules do not hold a charge, thus when sugar is dissolved in water, the solution does not conduct electricity, due to the lack of ions in the solution.

Therefore, depending on the ions that make up the compound, the substance would or would not conduct electricity.

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Question 1-19
Hunter-Best [27]

The experiment involving the determination of the number of ice cubes required to keep the temperature of the glass under 15 degrees Celcius, the following things have to be kept in mid:

  • The<u> temperature</u> of the surroundings
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  • The <u>number of ice cubes </u>added to the water in the glass

In order to keep into consideration the changing environmental temperatures (which is a variable in the experiment), the experiment had to be conducted daily to get <u><em>accurate results </em></u>keeping into consideration all the factors.

brainly.com/question/11256472

7 0
2 years ago
42.9 liters of hydrogen are collected over water at 76.0 °C and has a pressure of 294 kPa. What would the pressure (in kPa) of t
BabaBlast [244]

Answer:

494.1 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 294 kPa

P2 = ?

V1 = 42.9 liters

V2 = 22.8 liters

T1 = 76.0°C = 76 + 273 = 349K

T2 = 38.7°C = 38.7 + 273 = 311.7K

294 × 42.9/349 = P2 × 22.8/311.7

12612.6/349 = 22.8 P2/311.7

36.14 = 22.8P2/311.7

Cross multiply

36.14 × 311.7 = 22.8P2

11264.605 = 22.8P2

P2 = 11264.605 ÷ 22.8

P2 = 494.1 kPa

3 0
3 years ago
.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?
kipiarov [429]
H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄            +           2KOH ⇒ K₂SO₄ + 2H₂O
1mol                :           2mol
0,399mol         :           0,234mol
                                    limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
V = 0,950L + 0,900L = 1,85L

Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
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3 years ago
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