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3 years ago

Cyanite (Al2SiO5), quartz (SiO2), and leucite (KAlSi2O6) may be grouped together because they all contain ?

1 answer:

6 0

The answer is silicon and oxygen. They may be grouped under the largest class of mineral known as silicates. They contain oxygen and silicon which are the two most abundant elements in the earth's crust and it is estimated that they comprise more than 90 percent of the earth's hard outer layer.

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A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r

vampirchik [111]

We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)

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3 years ago

Which of the following is the best explanation for the fact that crushing increases the rate at which a solid dissolves in a liq

Oksana_A [137]

It’s D cause i know I know I know I know

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3 years ago

Most reactive O Be O Mg O Ba o Ca Which one is the most reactive

katen-ka-za [31]

Answer:

Explanation:

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3 years ago

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Aleks [24]

Answer:

10CO²+4H²O

Explanation:

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2 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 = E x r²

q = 1250 N/C x 0.205m²

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 = 5.84 x 10⁻⁹ C

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

3 0

3 years ago