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Ad libitum [116K]
3 years ago
14

A thin metallic spherical shell of radius 41.6 cm has a total charge of 8.55 μC uniformly distributed on it. At the center of th

e shell is placed a point charge of 4.43 μC . What is the magnitude of the electric field at a distance of 17.9 cm from the center of the spherical shell?
Chemistry
1 answer:
Paul [167]3 years ago
5 0

Answer:

E = 1.2443*10⁶ N/C

Explanation:

R = 41.6 cm = 0.416 m

Q₁ = 8.55 μC = 8.55*10⁻⁶C

Q₀ = 4.43 μC = 4.43*10⁻⁶C

r = 17.9 cm = 0.179 m

K = 9*10⁹ N*m²/C²

Since r < R  we can apply Gauss's Law as follows

E = K*Q₀ / r²

⇒  E = (9*10⁹ N*m²/C²)*(4.43*10⁻⁶C) / (0.179 m)²

⇒  E = 1.2443*10⁶ N/C

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 How much energy is needed to raise the temperature of 125g of water from 25.0oC to 35.0oC?  The specific heat of water is 4.184
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Hello!

To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.

In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.

Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.

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What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..
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Molar mass of NH_3

\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

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\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

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\\ \sf\longmapsto N_2+3H_2=2NH_3

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Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

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\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

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\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

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For Nitrogen

\\ \sf\longmapsto 1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{23}molecules

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Convert 1.8 mols of Br to its mass:

1.8 mols Br × 79.9g per mol = 143.82g of Br.
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