The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:

Mole ratio of Na2S and Al(NO3)3 = 3:2
Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25
= 0.0125 mole
Equivalent mole of Na2S = 3/2 x 0.0125
= 0.0188 mole
Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2
= 0.094 L or 94 mL
More on stoichiometric calculations can be found here: brainly.com/question/8062886
No. of moles= mass given /mass acquired
.0079=x/238
238×.0079=x
1.88~2
therefore the no. of grams =2g
The electric potential due to ammonia at a point away along the axis of a dipole is 1.44
10^-5 V.
<u>Explanation:</u>
Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.
Given p = 1.47 D = 1.47
3.34
10^-30 = 4.90
10^-30.
V = 1 / (4π∈о)
(p cos(θ)) / (r^2)
where p is a permanent electric dipole,
∈ο is permittivity,
r is the radius from the axis of a dipole,
V is the electric potential.
V = 1 / (4
3.14
8.85
10^-12)
(4.90
10^-30
1) / (55.3
10^-9)^2
V = 1.44
10^-5 V.
Body temperature is the main one. blood ph is very important. and reflex reactions. if this was right please give me brainliest