Answer:
2PbSO4 → 2PbSO3 + O2
Explanation:
in original equation we notice that we have one extra oxygen, which we cannot form a O2 with, so by multiplying everything else by 2, we get 2 extra oxygen
Answer:
526g is the mass of this sample
Explanation:
To solve this question we must, as first, find the <em>molar mass </em>of Al₂(Cr₂O₇)₃ using the periodic table. The molar mass is defined as the mass of this compound per mole. With this value we can find the mass in 0.750 moles as follows:
<em>Molar mass Al₂(Cr₂O₇)₃</em>
2Al = 2*26.98g/mol = 53.96g/mol
6 Cr = 6*51.9961g/mol = 311.9766g/mol
21 O = 21*15.999g/mol = 335.979g/mol
53.96g/mol + 311.9766g/mol + 335.979g/mol
= 701.9156g/mol
The mass of 0.750 moles is:
0.750 moles * (701.9156g / mol) =
<h3>526g is the mass of this sample</h3>
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>
The answer is A
According to research I have done, pure solids and liquids are not included in the equilibrium constant expression. If the concentration of a reactant in aqueous solution is increased, the position of equilibrium will move in the direction which minimises the effect of this increase in concentration, by using the added component up, to decrese it's concentration again.
