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:

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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:
Partial pressure of nitrogen gas is 0.98 bar.
Explanation:
According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.




where,
= total pressure = 3.9 bar
= partial pressure of nitrogen gas
= partial pressure of oxygen gas
= partial pressure of argon gases
= Mole fraction of nitrogen gas = 0.25
= Mole fraction of oxygen gas = 0.65
= Mole fraction of argon gases = 0.10
Partial pressure of nitrogen gas :

Partial pressure of oxygen gas :

Partial pressure of argon gas :

ANSWER: The two skills that Wade will need to excel in the career of a Radiologist are:
1. Interpersonal skills to talk to patients and help them feel comfortable: This skill important because patients often are afraid of getting detected by a disease. Few patients also fear of getting tested by the heavy and complicated machineries. A radiologist's interpersonal skill will calm the patient and his attendant and will make them feel relieved.
2. Technical skills to know how to use the machinery and equipment needed for the job: Radiologist's primary duty is to operate the machineries in order to diagnose a patient correctly. A radiologist can not afford to be an amateur as he will either not be able to detect a disease or may raise a false alarm of a person having a disease.
Answer:
20 kJ/mol
Explanation:
From ∆G°= -RTlnK
But
Ag2SO4(s)<----------->2Ag+(aq) + SO4^2-(aq)
Ksp= [2Ag+]^2 [SO4^2-]
But Ag+ = 0.032M
Ksp= (2×0.032)^2 (0.032)
Ksp= 1.31072×10^-4
∆G°= -RTlnK
∆G°= -(8.314× 298×(-8.93976))= 20KJmol-1( to the nearest KJ)
Answer:
Fe
Explanation:
The electrical conductivity depends mainly on the type of chemical bonds between the atoms of a compound.
In the case of MgF2, FeCl3 and FeO3, these have the type of ionic bond. This means that in the atoms of the compound there is an electron transfer, to keep eight electrons in the outermost layer and thus resemble the electronic configuration of the inert gas closest to each of the two elements, due to this ions of opposite charges are formed that are held together by electrostatic forces. These types of compounds are good conductors of electricity, however, to have this property, they must be dissolved in water or molten.
In the case of Fe, however, the type of union between atoms is metallic. In this type of junction, valence electrons are quite free inside the metal, which makes it easy for them to move. For this reason, this compound will conduct electricity in a solid state.