Answer:
(a) 0.22 mol Cl₂ and 15.4g Cl₂
(b) 2.89.10⁻³ mol O₂ and 0.092g O₂
(c) 8 mol NaNO₃ and 680g NaNO₃
(d) 1,666 mol CO₂ and 73,333 g CO₂
(e) 18.87 CuCO₃ and 2,330g CuCO₃
Explanation:
In most stoichiometry problems there are a few steps that we always need to follow.
- Step 1: Write the balanced equation
- Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
- Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.
(a)
Step 1:
2 Na + Cl₂ ⇄ 2 NaCl
Step 2:
In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.
Step 3:
(b)
Step 1:
HgO ⇄ Hg + 0.5 O₂
Step 2:
<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.
Step 3:
(c)
Step 1:
NaNO₃ ⇄ NaNO₂ + 0.5 O₂
Step 2:
<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.
Step 3:
(d)
Step 1:
C + O₂ ⇄ CO₂
Step 2:
<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.
Step 3:
![[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2](https://tex.z-dn.net/?f=%5Btex%5D20.0kgC.%5Cfrac%7B1%2C000gC%7D%7B1kgC%7D%20.%5Cfrac%7B44gCO_%7B2%7D%7D%7B12gC%7D%20%3D73%2C333gCO_%7B2)
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(e)
Step 1:
CuCO₃ ⇄ CuO + CO₂
Step 2:
<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.
Step 3:
Answer:
10
Explanation:
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺]
Thus, from the formula, more the concentration of the hydrogen ions or more the acidic the solution is, the less is the pH value of the solution.
Thus, solution with pH = 3 will be more acidic than solution with pH =4
Thus, concentration of the [H⁺] when pH =3
3 = - log [H⁺]
[H⁺] = 10⁻³ M
For pH = 4, [H⁺] = 10⁻⁴ M
<u>hence, pH = 3 is 10 times more acidic than pH = 4</u>
Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:
<span>H2C=O---------H-OH </span>
<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>
<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>
<span>H2C=O + H2O <-----------------> H2C(OH)2</span>
Answer
- continuous removal of PH3
- adding more of P into the system
Explanation:
In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);
- The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:
heat+A⇌BΔH=+
- Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
- According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.
