"Endothermic" is the way among the following choices given that you would <span>classify the following chemical reaction, in terms of energy. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has come to your help.</span>
<span>Thyroid hormones T4, T3
</span><span>Cortisol
</span><span>Estrogen or testosterone
</span><span>Insulinlike growth factor-I (IGF-I)</span>
Answer:
18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Explanation:
Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.
The decomposition of mercury oxide is given by the chemical equation below:
2HgO ----> 2Hg + O₂
2 moles of HgO decomposes to produce 1 mole of Hg
2 moles of HgO has a mass of 433.2 g
433.2 g of HgO produces 216.6 g of Hg
18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg
Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury
Answer:
Options B and C
Explanation:
Let's take a look at the options and get our answer by way of elimination. The basic definition of a neutral solution is given as;
A neutral solution is a substance which is neither acid nor basic . it has a PH of 7. it will have equal amount of H+ AND OH- ions in it.
a) a neutral solution does not contain any H3O+ or OH- This is wrong because take water as an example, it is neutral but contains both ions.
b) a neutral solution contains [H2O] = [H3O+]. This option is correct cause it is in line with the definition above.
c) an acidic solution has [H3O⁺] > [OH⁻]. Acidic solutions are any solution that has a higher concentration of hydrogen ions than water. This option is correct.
d) a basic solution does not contain any H3O⁺. This option is wrong. Basic solutions are any solution that has a higher concentration of hydroxide ions than water. This means they contain H3O⁺ but [OH⁻] is greater.
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
