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3 years ago

3. 79,789kJ = (293k) (377kJ)

Chemistry

1 answer:

KatRina [158]3 years ago

5 0

the anwser would most likely be j=0

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open the phase diagram for co2 given in the introduction again. use the phase diagram for co2 in the interactive activity and de

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The phase diagram of CO2 has a melting curve that slopes up and to the right, in contrast to the phase diagram of water, which has a more conventional shape. It is impossible for liquid CO2 to exist at pressures lower than 5.11 atm because the triple point is 5.11 atm and 56.6 °C.

Due to the fact that ice is less thick than liquid water, the phase diagram of water has an odd melting point that drops with pressure. Carbon dioxide cannot exist as a liquid at atmospheric pressure, according to the phase diagram of the gas. Thus, gaseous carbon dioxide directly sublimes from solid carbon dioxide.

Learn more about solid carbon dioxide.

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1 year ago

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3 years ago

Find the mass of 1 atom of Chlorine in gram.

vazorg [7]

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3 years ago

What elements are found in salt?

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3 years ago

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

alina1380 [7]

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>

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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3 years ago