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Bingel [31]
3 years ago
7

I NEED HELP ASAP!! What is the difference between the experimental group and a control group?​

Chemistry
1 answer:
dexar [7]3 years ago
3 0

I believe the control group is what doesn't change in the experiment, and the experimental group is what is being tested / receives the treatment :)

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A space probe identifies a new element in a sample collected from an asteroid. Successive ionization energies (in attojoules per
melomori [17]
Ionization energy (IE) is the amount of energy required to remove an electron.

If you observe the IEs sequentially, there is a large gap between the 2nd and 3rd. This suggests it is difficult to remove more than 2 two electrons. Elements that lose two electrons to become more stable are found in the Group 2A (2 representing the number of electrons in the outermost valence shell).

4 0
3 years ago
Read 2 more answers
The ionic radius for Na+ is 0.097 ηm and for Cl- is 0.181 ηm, the absolute value of the charge for each ion is 1.6x10-19 C, ε_o=
Vanyuwa [196]

Answer:

B = (2.953 × 10⁻⁹⁵) N.m⁹

Explanation:

At equilibrium, where the distance between the two ions (ro) is the sum of their ionic radii, the force between the two ions is zero.

That is,

Fa + Fr = 0

Fa = - Fr

Fa = (|q₁q₂|)/(4πε₀r²)

Fr = -B/(r^n) but n = 9

Fr = -B/r⁹

(|q₁q₂|)/(4πε₀r²) = (B/r⁹)

|q₁| = |q₂| = (1.6 × 10⁻¹⁹) C

(1/4πε₀) = k = (8.99 × 10⁹) Nm²/C²

r = 0.097 + 0.181 = 0.278 nm = (2.78 × 10⁻¹⁰) m

(k|q₁q₂|)/(r²) = (B/r⁹)

(k × |q₁q₂|) = (B/r⁷)

B = (k × |q₁q₂| × r⁷)

B = [8.99 × 10⁹ × 1.6 × 10⁻¹⁹ × 1.6 × 10⁻¹⁹ × (2.78 × 10⁻¹⁰)⁷]

B = (2.953 × 10⁻⁹⁵) N.m⁹

6 0
3 years ago
PLEASE HELP 20 points
777dan777 [17]

  n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl

  NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl

  58.44g NaCl x .98 mol NaCl = 57.27g NaCl

Explanation:

hope you get it right :)

4 0
2 years ago
Give the symbol for the element with the following orbital diagram:
wel

Answer: 0%

Explanation:

You got a bad grade

5 0
3 years ago
A solution has a [OH-] of 1 × 10-9. What is the pOH of this solution?
zhannawk [14.2K]
POH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁹ ] 

pOH = 9

Answer C

hope this helps!
3 0
3 years ago
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