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gogolik [260]
4 years ago
10

John recently purchased a High Definition plasma television and needed to have it installed. The installers told John that for t

he best picture quality he should use cables with an all-copper core, rubber insulation, and 24k gold contacts. The electrical signal travels down the copper portion of the cable while the rubber insulates the cable from a loss of electrical current and heat. Which physical property explains the difference in ability of copper and rubber to transmit electricity?
ductility
density
conductivity
Chemistry
2 answers:
ioda4 years ago
8 0
Ductility - a materials ability to stretch, ie if you pull it apart does it stretch to a wire.

density - ratio of volume to mass

conductivity - materials ability to conduct a current.

hopefully with these definitions you can figure out the answer.
svlad2 [7]4 years ago
3 0

the answer's conductivity i just did it lol

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Which of the following statements are correct? More than one answer may be correct.
garri49 [273]

Answer:

E, only one statement is correct

Explanation:

Absorption is favoured by heating the solution containing the colored impurity.

Pleaseee! Mark me brainliest

4 0
2 years ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
1.24 moles of magnesium arsenate are dissolved in 1.74 kg of solution. Calculate the molality of the solution.
aleksklad [387]

Answer:

Molality of the solution = 0.7294 M

Explanation:

Given:

Number of magnesium arsenate = 1.24 moles

Mass of solution = 1.74 kg

Find:

Molality of the solution

Computation:

Molality of the solution = Mole of solute / Mass of solution = 1.74 kg

Molality of the solution = 1.24 / 1.7

Molality of the solution = 0.7294 M

6 0
3 years ago
Is diphenylamine a solid, liquid, or a gas at room temperature
goblinko [34]

Answer:

Solid

Explanation:

Diphenylamine has a melting point of 127.4 F or 53 C so at room temperature ~70 F or 21 C its a solid.

3 0
3 years ago
How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
nevsk [136]

Answer:

17.5 g

Explanation:

Given data

  • Mass of solution to be prepared: 50.0 grams
  • Concentration of the salt solution: 35.0%

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl

4 0
3 years ago
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