Answer:
ΔH = 1270kJ
Explanation:
Let's consider the following thermochemical equation.
B(s) + 3/2 O₂(g) → B₂O₃(s), ΔH = −1270kJ
Since ΔH < 0, this reaction is exothermic. According to the Lavoisier-Laplace law, if we reverse a reaction, its enthalpy will have the same value and opposite sign than the direct reaction.
B₂O₃(s) → B(s) + 3/2 O₂(g), ΔH = 1270kJ
For the inverse reaction, ΔH > 0, which means that it is endothermic.
Answer:
= 1 X 10⁻⁻¹²M
Explanation:
At 25°C & 1atm [H⁺][OH⁻] = 1 x 10⁻¹⁴ => [H⁺] = 1 X 10⁻¹⁴/[OH⁻] = 1 X 10⁻¹⁴/1 X 10⁻²
= 1 X 10⁻⁻¹²M
Answer:
Explanation:
Edit the reaction by drawing all steps in the appropriate boxes and connecting them with reaction arrows. Add charges where needed. Electron flow arrows should start on the electron(s) of an atom or a bond and should end on an atom, bond, or location where a new bond should be created. Include all free radicals by right-clicking on an atom on the canvas and then using the Atom properties to select the monovalent radical.
Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Molarity is defined as number of moles of solute in 1 L of solution.
Here, 0.1025 g of Cu is reacted with 35 mL of HNO_{3} to produced Cu^{2+} ions.
The balanced reaction will be as follows:
Cu+3HNO_{3}\rightarrow Cu(NO_{3})_{2}+NO_{2}+H_{2}O
From the above reaction, 1 mole of Cu produces 1 mole of Cu^{2+}, convert the mass of Cu into number of moles as follows:
n=\frac{m}{M}
molar mass of Cu is 63.55 g/mol thus,
n=\frac{0.1025 g}{63.55 g/mol}=0.0016 mol
Now, total molarity of solution, after addition of water is 200 mL or 0.2 L can be calculated as follows:
M=\frac{n}{V}=\frac{0.0016 mol}{0.2 L}=0.008 mol/L=0.008 M
Thus, molarity of Cu^{2+} is 0.008 M.