Answer:
0.400 moles
Explanation:
Well, you know that one mole of ammonium chloride,
NH 4 Cl
, dissociates completely in aqueous solution to form one mole of ammonium cations,
NH + 4
, and one mole of chloride anions, Cl
−
.
Answer:
Mode
Explanation:
The mode is the number that appears most frequently in a data set. A set of numbers may have one mode, more than one mode, or no mode at all. Other popular measures of central tendency include the mean, or the average (mean) of a set, and the median, the middle value in a set.
Answer:- The gas needs to be transferred to a container with a volume of 11.2 L.
Solution:- From Boyle's law. "At constant temperature, Volume is inversely proportional to the pressure."
It means, the volume is decreased if the pressure is increased and vice versa.
Here, the Pressure is decreasing from 537 torr to 255 torr. So, the volume must increase and calculated by using the equation:

Where,
is initial pressure and
is final pressure. Similarly,
is initial volume and
is final volume.
Let's plug in the values in the equation:
(537 torr)(5.30 L) = (255 torr)(
)

= 11.2 L
So, the new volume of the container needs to be 11.2 L.
Answer:
1. Alkali metals (group 1)
2. halogens (Group 17)
3. noble gasses (group 18)
Explanation:
1. alkali metals only have one valence electron meaning that they really want to lose that one valence electron to get a full octet.
2. halogens have 7 valence electrons meaning that they just need to gain 1 to get a full octet.
3. Nobel gasses already have a full octet meaning that they don't want to react. (atoms only react to get a full octet)
I hope this helps. Let me know if anything is unclear.
Answer:
7.43 × 10²⁴ m⁻³
Explanation:
Data provided in the question:
Conductivity of a semiconductor specimen, σ = 2.8 × 10⁴ (Ω-m)⁻¹
Electron concentration, n = 2.9 × 10²² m⁻³
Electron mobility,
= 0.14 m²/V-s
Hole mobility,
= 0.023 m²/V-s
Now,
σ = 
or
σ = 
here,
q is the charge on electron = 1.6 × 10⁻¹⁹ C
p is the hole density
thus,
2.8 × 10⁴ = 1.6 × 10⁻¹⁹( 2.9 × 10²² × 0.14 + p × 0.023 )
or
1.75 × 10²³ = 0.406 × 10²² + 0.023p
or
17.094 × 10²² = 0.023p
or
p = 743.217 × 10²²
or
p = 7.43 × 10²⁴ m⁻³