Answer:
13.2 g of gold
Explanation:
We'll begin by converting 5.25 L to ft³.
This can be obtained as follow:
Recall:
1 L = 0.0353 ft³
Therefore,
5.25 L = 5.25 × 0.0353
5.25 L = 1.85×10¯¹ ft³
From the question given above,
2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.
Therefore, Xg of gold will be present in 1 ft³ of soil i.e
Xg of gold = 2.45/1.85×10¯¹
Xg of gold = 13.2 g
Therefore, 13.2 g of gold is present in 1 ft³ of the soil.
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
Answer:
Kc = 3.90
Explanation:
CO reacts with
to form
and
. balanced reaction is:

No. of moles of CO = 0.800 mol
No. of moles of
= 2.40 mol
Volume = 8.00 L
Concentration = 
Concentration of CO = 
Concentration of
= 

Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium
= 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium
= 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)

Answer:
b. a new gas is observed in the form of bubbles
Explanation:
The only statement that was evidence of a chemical reaction is a new gas is observed in the form of bubbles Any time a new substance is created, a chemical reaction has taken place.