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Sergeu [11.5K]
4 years ago
8

1. In the models, what particles do the colored and plain sesame seeds represent? What particles do the poppy seeds represent

Chemistry
2 answers:
Veronika [31]4 years ago
6 0
Ok the answers to the hole .doc is 
1. Neutrons, Protons, and Electrons
2. N<span>ucleus
3. N</span>eutrons and Protons
4. Electrons
5. Because they represent different things (I would put this in your own words)
enyata [817]4 years ago
4 0
<h2>Answer 1) : The colored and plain sesame seeds represent the subatomic particles neutrons and protons. </h2><h3>Explanation :</h3>

The attached model of the atom has subatomic particles as neutrons and protons; which are present in the center of the nucleus of the atom. Where neutrons are neutrally charged particles in the nucleus and protons are positively charged particles present in the nucleus of the atom.

<h2>Answer 2) The poppy seeds represent the electrons.</h2><h3>Explanation :</h3>

The poppy seeds used in this atomic model represents the revolving electrons around the the nucleus of the atom in orbits. These are negatively charges subatomic particles which revolves around the nucleus.

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Alex17521 [72]
Earth's atmosphere is 78% nitrogen, 21% oxygen, 0.9% argon, and 0.03% carbon dioxide with very small percentages of other elements. Our atmosphere also contains water vapor. In addition, Earth's atmosphere contains traces of dust particles, pollen, plant grains and other solid particles.
5 0
4 years ago
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Which sublevels are full when an atom has a noble gas configuration?
NemiM [27]
<h3>Answer:</h3>

s-sub-level and p-sub-level

<h3>Explanation:</h3>
  • Group 18 elements in the periodic table are known as noble gases.
  • They have an octet configuration with their outermost s and p-sub-levels full, which makes them inert or nonreactive.
  • The s-sub-level takes up a maximum of 2 electrons while the p-sub-level holds a maximum of 6 electrons.
  • They include, argon, neon, xenon, krypton, etc.
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  • For example; the electron configuration of Ne is 1s²2s²2p⁶, with full s and p sub-levels.
6 0
4 years ago
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a faucet is leaking water at a speed of 5 drops per minute. If 1 ml is equivalent to 10 drops, how many liters of water will be
nikdorinn [45]

Answer: 0.72 litres of water is wasted in one day.

Explanation:

First you need to find out how many minutes are in a day. Do this by multiplying the number of minutes in an hour (60) by the number of hours in a day (24). 24 x 60 = 1440. If the faucet is dripping at 5 drops per minute, then multiply 5 by the number of minutes in a day (1440) to see how many drops drip in one day. 5 x 1440 = 7200. Now we need to figure out how many mL fo water that is. if 10 drops is 1 mL, then we need to divide the total number of drops (7200) by 10. 7200 divided by 10 is 720. That means 720 mL of water is dripping per day. Finally, we must convert mL to litres. There are 1000 mL in one litre, so divide 720 by 1000. The final answer is 0.72

6 0
3 years ago
For many purposes we can treat butane (C4H10) as an ideal gas at temperatures above
masha68 [24]

Answer:

A. Yes

B. –176 °C

Explanation:

A. Yes

B. Determination of the new temperature of the gas.

Let the initial pressure be P

From the question given above, the following data were obtained.

Initial pressure (P1) = P

Initial temperature (T1) = 19 °C

Final pressure (P2) = ⅓ P1 = ⅓P = P/3

Final temperature (T2) =?

Next, we shall convert 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273

Initial temperature (T1) = 292 K

Since the volume is constant, we can obtain the new temperature of the gas as illustrated below:

Initial pressure (P1) = P

Initial temperature (T1) = 292 K

Final pressure (P2) = P/3

Final temperature (T2) =?

P1/T1 = P2/T2

P/292 = P/3 /T2

P/292 = P/3T2

Cross multiply

P × 3T2 = 292 × P

Divide both side by P

3T2 = (292 × P)/P

3T2 = 292

Divide both side by 3

T2 = 292/3

T2 = 97.33 ≈ 97 K

Finally, we shall convert 97 K to celcius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 97

T(°C) = 97 – 273

T(°C) = –176 °C

Thus, the new temperature of the gas is –176 °C.

6 0
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