The amount of W(OH)2 needed would be 448.126 g
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
W(OH)2 + 2 HCl → WCl2 + 2 H2O
The mole ratio of W(OH)2 to HCl is 1:2
Mole of 150g HCl = 150/36.461
= 4.11 moles
Equivalent mole of W(OH)2 = 4.11/2
= 2.06 moles
Mass of 2.06 moles W(OH)2 = 2.06 x 217.855
= 448.188g
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Potassium sulfide, also
known as dipotassium monosulfide, consists of two potassium ions bonded to a
sulfide atom, rendering the chemical formula K2S.<span>Rarely
found in nature due to its high reactivity with water, potassium sulfide is
refined from the more common potassium sulfate (K2SO4) and is used in many
industries</span>
Answer:
false
Explanation:
the energy causes the bonds to become looser
There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
