We are given with a vinegar with a hydrogen ion concentration of 0.00010 m. We are asked to express this concentration in scientific notation. The answer when expressed in scientific notation is 1x10^-4 m or molality. Answer is <span>1x10^-4 m. </span>
The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:

=

Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).

=

∴ by substituting the known values, ⇒ (103 kPa) ÷ (25 °C) = (P₂) ÷ (50 °C)
⇒ P₂ = (4.12 kPa · °C) (50 °C)
=
206 kPa
Thus the pressure of the gas since the temperature was raised from 25 °C to 50 °C is
206 kPa
Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.
Answer:
Option B. 2Mg(s) + O2 (g) —> 2MgO (s)
Explanation:
From the question given above,
We were told that:
2 solid Mg atoms bond with O2 gas to produce solid MgO.
This can be represented by an equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
Next, we shall balance the above equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
There are 2 atoms of Mg on the left side and 1 atom on the right side. It can be balance by putting 2 in front of MgO as shown below:
2Mg(s) + O2 (g) —> 2MgO (s)
Now, the equation is balanced.
Answer:
(a) The anode electrode which comprises the zinc electrode being placed in a water solution with low oxygen concentration.
(b) Cathodic reaction is:
⇒ 
Anodic reaction is:
⇒
Explanation:
In the given problem, we have an oxygen-concentration cell consisting of two zinc electrodes. One is immersed in a water solution with a low oxygen concentration and the other in a water solution with a high oxygen concentration. The zinc electrodes are connected by an external copper wire.
(a) Which electrode will corrode?
The electrode that will corrode is the anode electrode which comprises the zinc electrode being placed in a water solution with low oxygen concentration.
(b) Write half-cell reactions for the anodic reaction and the cathodic reaction.
Cathodic reaction is:
⇒ 
Anodic reaction is:
⇒