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Nadusha1986 [10]
1 year ago
7

A rod, X has a positive charge of 8. An otherwise identical rod, Y has a negative charge of 4. The rods are touched together, an

d then separated.
1.When they touch, what particles move between them?

2.Did the particles move from "X" or "Y" or from "Y" to "X"?
Chemistry
1 answer:
gtnhenbr [62]1 year ago
4 0

Answer:

1.  electrons

2. From "Y" to "X"

Explanation:

1. Electrons move between the rod since the electrons are the only charge carriers which are free to move.

2. The particles move from from "Y" to "X"  since the electrons are the only charge carriers which are free to move. The positive charge on rod x is due to a deficit of electrons while the negative charge on rod Y is due to the excess of electrons. When the rods come together, the electrons move from "Y" to "X" since the electrons are the only charge carriers which are free to move.

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The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

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Since, 1 gal= 3785.41 mL

Thus, 2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL

Density of water is 1 g/mL thus, mass of water will be 8.66\times 10^{7}g.

Since, 1 grams of chlorine →10^{6} grams of water.

1 g of water →10^{-6} g of chlorine and,

8.66\times 10^{7}g of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

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2 years ago
A block of aluminum weighing 140 g is cooled from 98.4°C to 62.2°C with the release of 1080 joules of heat. From this data, calc
GrogVix [38]

<u>Answer:</u>

Specific heat of a substance is the value that describe how the added heat energy of substance has the impact on its temperature.

Unit is <em>(\frac {J}{Kg.K})</em>

<em>C = Q/m. ∆T</em>

<em>C – Specific heat (\frac {J}{Kg.K})</em>

<em>Q- heat energy (J)</em>

<em>M – Mass (Kg)</em>

<em>∆T- change in temperature (K) </em>

<u>Explanation:</u>

<em>Given data:</em>

<em>M= 140 g = 0.14 Kg</em>

<em>Q – 1080 Joules.</em>

<em>∆T – 98.4 – 62.2 = 36.2</em>

Substituting  the given data in Equation

<em>Specific heat of Aluminium  = \frac {1080}{(0.14 \times 36.2)} = 213.10 (\frac {J}{Kg.K})</em>

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