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irina [24]
4 years ago
12

A scientist performs an experiment on an unknown element. She finds that the element forms an ionic bond with beryllium (Be) but

not with lithium (Li). She concludes that the unknown element must belong in group 2 of the periodic table. State whether or not you think this result supports her conclusion, and why
Physics
2 answers:
Bumek [7]4 years ago
4 0

To fill up all of the valence electrons we must have 8 electrons total to have an ionic bond.  In this case Lithium is in group 1 and so only has 1 valence electron and so it can bond with elements from group 17, since 7 + 1 = 8.  However, this unknown element did bond with beryllium, and beryllium is in group 2 so it must bond with an atom that has 6 valence electrons.  The elements that have 6 valence electrons are in group 16 not group 2.  So the result does not support her conclusion.

aivan3 [116]4 years ago
3 0

Answer: The result do not support her conclusion as group 2 elements are metals.

Explanation:

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and forms a positively charged ion called as cation. The element which accepts the electrons is known as electronegative element or non metal and forms a negatively charged ion called as anion.

As both Lithium (Li) and berrylium (Be) are metals, they would combine with a non metal to form ionic bond. Thus the unknown element must be a non metal and should belong to group 14 , 15 , 16 or 17 and not to group 2, which are metals.

Thus The result do not support her conclusion as group 2 elements are metals.

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Lady_Fox [76]
Based on the given statement above, the correct answer would be FALSE. It is not true that range of motion is the distance an object can travel when separated from another object because range of motion or ROM is the distance--linear or angular--<span>that a movable object may normally travel while properly ATTACHED (not separated) to another. Hope this answer helps.</span>
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3 years ago
a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second
lawyer [7]

Answer:

a) Yes, there are 10^8 shakes in a second (1 s \frac{1 shake}{10^{-8}s}=10^8 shake) in a year there are 31,536,000 seconds... that is 3.1536x10^7 (1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7)

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s

That is about 2 and half hours.

3 0
4 years ago
6. What do the key results indicate?
slega [8]

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7 0
4 years ago
Read 2 more answers
A hot-air balloon is rising upward with a constant speed of 2.51 m/s. When the balloon is 3.16 m above the ground, the balloonis
icang [17]

Answer:

  t = 1.099 s

Explanation:

given,

constant speed = 2.51 m/s

height of balloon above ground = 3.16 m

time elapsed before it hit the ground = ?

Applying equation of motion to the compass

y = u t + \dfrac{1}{2}at^2

-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2

4.9 t^2 - 2.51 t - 3.16 = 0

using quadratic formula to solve the equation

t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}

  t = 1.099 s, -0.586 s

hence, the time elapses before the compass hit the ground is equal to 1.099 s.

8 0
4 years ago
A young 34-kg ice hockey goalie, originally at rest, catches a 0.145-kg hockey puck slapped at him at a speed of 34.5 m/s. In th
irinina [24]

The concept that we need to give solution to this problem is collision equation given by momentum conservation,

Our values are,

m_2 = 0.145 kg\\u_2 = 34.5 m/s\\m_1 = 34 kg\\u_1 = 0

Then,

Part A) We can here note that the velocity for the puck is zero (there is not a velocity in that direction)

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V_{goalie} = \frac{34-0.145}{34+0.145}(0)+\frac{2*0.145}{34+0.145}(34.5)

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Part B ) We apply the same solution but know we note that in the collision for the Goalie the velocity is zero.

V_{puck} = \frac{m_1-m_2}{m_1+m_2}V_{02} + \frac{2m_2}{m_1+m_2}V_{01}

V_{puck} = \frac{34-0.145}{34+0.145}(34.5)+\frac{2*0.145}{34+0.145}(0)

V_{puck} = 34.20m/s

7 0
3 years ago
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