Answer:
0.2 mL stock solution, 0.8 solvent, 0.1 mL first solution and 0.9 solvent
Explanation:
The final volume for fist solution is 1 mL and concentration must will be 1/5, then 1 mL/5=0.2 mL. For complete the 1 mL add the missing solvent volume 1 mL-0.2 mL=0.8 mL. For second solution, assuming final volume is 1 mL, and concentration 1/10, then we have 1 mL /10=0.1 mL solution 1/5. Completing volume, 1 mL-0.1 mL= 0.9 mL solvent.
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
I believe the answer is D
Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is
.
Explanation:
When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.
For example, chemical equation for oxidation of methane is as follows.

Number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
To balance this equation, multiply
by 2 on reactant side. Also, multiply
by 2 on product side. Hence, the equation can be rewritten as follows.

Now, the number of atoms present on reactant side are as follows.
Number of atoms present on product side are as follows.
Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.
Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is
.
I think the correct answer would be D. The reaction that involves an acid and a covalent base would be the reaction of sulfuric acid and water or H2SO4 + 2H2O → 2H3O+ + SO42– . The acid would be H2SO4 and the covalent base would be H2O since it is being held by covalent bonds and when in solution it will have equal amounts of OH- and H+ ions.