Answer:
a) Ag(NH₃)₂⁺, Cl⁻.
b) NH₃.
c) AgCl.
Explanation:
Based on LeChatelier's law, a system in chemistry can change responding to a disturbance of concentration, temperature, etc. in order to restore a new state.
In the reaction:
AgCl(s) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺(aq) + Cl⁻(aq)
When reactants are added, the system will produce more products restoring the equilibrium and vice versa. A reactant in solid state doesn't take part in the equilibrium, thus:
a) Ag(NH₃)₂⁺, Cl⁻. The addition of products will shift the equilibrium to the left
b) NH₃. The addition of reactant will shift the equilibrium to the right.
c) As AgCl is in solid phase, will not shift the equilibrium in either direction.
<span>así que te está diciendo que</span>
Answer:
Half life of phosphorous-32 = 14 days
Explanation:
Given data:
Total mass of phosphorous-32 = 2.0 g
After 42 days mass left = 0.25 g
Half life of phosphorous-32 = ?
Solution:
First of all we will calculate the number of half lives passed.
At time zero = 2.0 g
At first half life = 2.0 g/2 = 1.0 g
At 2nd half life = 1.0 g/2 = 0.5 g
At 3rd half life = 0.5 g/2 = 0.25 g
Half life:
Half life = T elapsed / half lives
Half life = 42 days/ 3
Half life = 14 days
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
