Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
Due to energy being a reactant instead of a product, the process is endothermic. The system must absorb a quantity of energy before it can react, so it must be an endothermic system.
For the conversions
I will start with pressure
1atm=101.3kPa
x =700kPa
x=700kPa/101.3kPa
x=6.91atm
Temperature
273K+30.00C
303K
Volume
1L=1000ml
x =50ml
x=0.05L
PV=nRT
6.91*0.05=n*0.08206*303
0.3455=24.86418n
0.3455/24.86418=n
0.0138=n
number of moles = 0.0138moles
Note: 0.08206 is the gas constant in this case
Answer:
96.5 g/ml
Explanation:
If 5g is 19.3 then 25g is 19.3x5 which is 96.5 g/ml
Answer:
105.8 g of Na would be required
Explanation:
Let's think the reaction:
2Na(s) + Cl₂(g) → 2NaCl (s)
1 mol of chlorine reacts with 2 moles of sodium
Then, 2.3 moles of Cl₂ would react with (2.3 .2) / 1 = 4.6 moles
Let's determine the mass of them.
4.6 mol . 23 g/mol = 105.8 g
