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GarryVolchara [31]
3 years ago
14

How many excess electrons must be added to an isolated spherical conductor 41.0 cmcm in diameter to produce an electric field of

magnitude 1250 N/CN/C just outside the surface?
Chemistry
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

3.65 x 10¹⁰ electrons

Explanation:

we'll apply the following equation for electric field of a point charge on a spherical conductor

E = k \frac{q}{r^{2} }

where E is the electric field

k is a constant of the value 8.99 x 10⁹ Nm²/C²

r is the radius of the spherical conductor

q is the total charge in the sphere

Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)

Electrical field E = 1250 N/C

we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field

E = k \frac{q}{r^{2} }

q = <u>E x r²</u>

        k

q =  <u>1250 N/C x 0.205m</u>²

       8.99 x 10⁹ Nm²/C²

q =   5.84 x 10⁻⁹ C

this is the total charge in the sphere

To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)

n = \frac{q}{e}

n = <u>5.84 x 10⁻⁹ C </u>

       1.6 x 10⁻¹⁹C

n = 3.65 x 10¹⁰ electrons

Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons

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kondor19780726 [428]

Answer:

Here's what I get

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A substance with ρ > 1.36 g/cm⁻³ will sink in corn syrup

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Read 2 more answers
When 20.0 g of KI are dissolved in 50.0 mL of distilled water in a calorimeter, the temperature drops from 24.0 °C to 19.0 °C. C
Reil [10]
<h2>Answer:</h2>

<em>8.67kJ/mol</em>

<h2>Explanations</h2>

The formula for calculating the amount of heat absorbed by the water is given as:

\begin{gathered} q=mc\triangle t \\ q=50\times4.18\frac{J}{g^oC}\times(19-24) \\ q=50\times4.18\times(-5) \\ q=-1045Joules \\ q=-1.045kJ \end{gathered}

Determine the moles of KI

\begin{gathered} moles\text{ of KI}=\frac{mass\text{ of KI}}{molar\text{ mass of KI}} \\ moles\text{ of KI}=\frac{20g}{166g\text{/mol}} \\ moles\text{ of KI}=0.1205moles \end{gathered}

Since heat is lost, hence the enthalpy change of the solution will be negative that is:

\begin{gathered} \triangle H=-q \\ \triangle H=-(-1.045kJ) \\ \triangle H=1.045kJ \end{gathered}

Determine the enthalpy of solution in kJ•mol-1

\begin{gathered} \triangle H_{diss}=\frac{1.045kJ}{0.1205mole} \\ \triangle H_{diss}\approx8.67kJmol^{-1} \end{gathered}

Hence the enthalpy of solution in kJ•mol-1 for KI is 8.67kJ/mol

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