Answer:
This is based on most likely so it's A.Fe
A)Cu2O(s) + C(s) ———> 2Cu(s)+ CO(g)
B)H2(g) +Cl2(g)———> 2HCl(g)
Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
The solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L in water at 25°C.
At a specific solution temperature, a solid salt compound can entirely dissolve in pure water up to a predetermined molar solubility limit. The dissociation stoichiometry ensures that the molarities of the constituent ions are proportionate to one another. The saturable nature of the solution causes them to also coexist in a solubility equilibrium with the solid component. At this temperature, a solubility product constant Ksp is calculated using the solubility product of their molarity values.
Lead (II) fluoride has the following solubility equilibrium for its saturated solution:
⇄ 
![K_s_p = [Pb^2^+][F^-]^2](https://tex.z-dn.net/?f=K_s_p%20%3D%20%5BPb%5E2%5E%2B%5D%5BF%5E-%5D%5E2)
This compound dissociates in a 1:2 ratio of ions. For the compound dissolved in pure water, the Ksp is expressed in terms of the molar solubility "x" as:
Here,
× 
4.1 × 10⁻⁸ = 4 x³
x³ = 1.025 × 10⁻⁸
x³ = 10.25 × 10⁻⁹
x = 2.17 × 10⁻³ g/L
Therefore, the solubility of Lead(II)Fluoride is 2.17 × 10⁻³ g/L.
Learn more about solubility here:
brainly.com/question/23946616
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