Answer:
Volume = 3.86 ml (Approx)
Explanation:
Given:
Density of cadmium = 8.65 g/ml
Mass of pure object = 33.4 g
Find:
Volume pure cadmium
Computation:
Volume = Mass / Density
Volume = 33.4 / 8.65
Volume = 3.86 ml (Approx)
The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.
<h3>What is the dissociation enthalpy?</h3>
Given that;
H-H bond energy = 435 kJ mol^-1
H-Cl bond energy = 431 kJ mol^-1
ΔHfO of HCL(g) = -92kJ mol^-1
Bond dissociation enthalpy of the Cl-Cl bond = x
-92 = 435 + 431 + x
x = -92 - (435 + 431)
x = -958 kJ mol^-1
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The total mass of the products is 10.76 g + 204.44 g = 215.20 g.
The masses of all the reactants but one are known so,
215.20 g - 120.00 g - 8.15 g - 75.00 g = 12.05 g
12.05 g is the mass of the unweighed barium nitrate.
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:
64J of energy must have been released.
Explanation:
Step 1: Data given
One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.
The product contains 435 J of chemical energy.
Step 2:
Since the energy is conserved
Sum of energy of Reactants = Energy of Products
Sum of energy of Reactants = 346 J + 153 J = 499 J
The energy of the product = 435 J
435 < 499
This means energy must have been lost as heat.
Step 3: Calculate heat released
499 J - 435 J = 64 J
64J of energy must have been released.