Answer:
disposal of radioactive waste
Explanation:
if too much is released, it can wipe out large parts of the country
The PRODUCT is found on the right side of the arrow in a chemical reaction.
Answer:
38 elements or Element 31: Gallium Element 32: Germanium Element 33: Arsenic Element 34: Selenium Element 35: Bromine Element .
Explanation:
How many transition metals are there?
The 38 elements in groups 3 through 12 of the periodic table are called "transition metals". As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat.
Answer:
Approximately 
.
Explanation:
The gallium here is likely to be produced from a 
solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?
Note the Roman Numeral "
" next to 
. This numeral indicates that the oxidation state of the gallium in this solution is equal to 
. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of 
.
As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium solution.
How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.
.
.
Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.
.
It takes three moles of electrons to deposit one mole of gallium atoms . As a result, 
of electrons would deposit 
of gallium atoms .
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required
