Cr2(CO3)3 is Chromium (III) Carbonate or Chromic Carbonate.
It is composed of Chromium, Carbon, and Oxygen. Chromium has 2 atoms, Carbon has 3 atoms, and Oxygen has 9 atoms.
This compound is composed of 36.6% Chromium, 12.7% Carbon, and 50.7% Oxygen.
Answer:
a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) 3.14g must be added
Explanation:
a) For the reaction:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)
As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)
To balance hydrogens, the other coefficients are:
Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:
6.5 g × (1mol / 315.48g) =<em> 0.0206moles of Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:
0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:
0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>
Answer: 581 gmol
0.581 kmol
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
1. The conversion for mol to gmol
1 mol = 1 gmol
581 mol= 
2. The conversion for mol to kmol
1 mol = 0.001 kmol
581 mol= 
3. The conversion for mol to lbmol
1 mol = 
581 mol= 
100% find the gfm of both sides then divide
590 mL = 590 cm³= 0,59 dm³
C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________
M KNO₃ = 39g+14g+16g×3 = 101 g/mol
1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃
:)
