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Answer:
0.200L of 0.300 Al(OH)₃ are necessaries
Explanation:
A diprotic acid, H₂X, reacts with aluminium (III) hydroxide, Al(OH)₃ as follows:
3 H₂X + 2 Al(OH)₃ → 6H₂O + Al₂X₃
To solve this question we must find the moles of the H₂X, using the balanced reaction we can find the moles of Al(OH)₃ and its volumen knowing its molar concentration is 0.300M:
<em>Moles H₂X:</em>
0.300L * (0.300mol / L) = 0.0900moles H₂X
<em>Moles Al(OH)₃:</em>
0.0900moles H₂X * (2mol Al(OH)₃ / 3mol H₂X) = 0.0600 moles Al(OH)₃
<em>Volume 0.300M Al(OH)₃:</em>
0.0600 moles Al(OH)₃ * (1L / 0.300moles) =
<h3>0.200L of 0.300 Al(OH)₃ are necessaries</h3>
Answer:
884.56 torr
Explanation:
Formula: 
P = Pressure
V = Volume
T = Temperature in kelvin (Celsius + 273.15)
P = 884.56169
Answer:
Answer in explanation
Explanation:
a. Boron , element 5
Helium has 2 electrons, add to the other 3 to give 5.
Other group members are : Aluminum Al, Gallium Ga, Indium In , Thallium Tl and Nihonium Nh
b. Sulphur, element 16
Neon is 10 , add other 6 electrons to make 16
Other group members are: Oxygen O, selenium Se , Tellurium Te and Polonium Po
c. Lanthanum, element 57
Xenon is 54, add the other 3 electrons to give 57.
Other elements in group : Scandium Sc , Yttrium Y , Actinium Ac, Lutetium Lu and/or Lawrencium Lr
