We can calculate for temperature by assuming the equation
for ideal gas law:
P V = n R T
Where,
P = pressure = 1.80 atm
V = volume = 18.2 L
n = number of moles = 1.20 moles
R = gas constant = 0.08205746 L atm / mol K
Substituting to the given equation:
T = P V / n R
T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm /
mol K)
T = 332.70 K
We can convert K unit to ˚C unit by subtracting 273.15
to Kelvin, therefore
T = 59.55 ˚<span>C</span>
The molecular model is physical model of the atomic system used to <span>represent </span>molecules and their processes. It is used to <span>understand </span>chemistry<span> and generating and testing </span>hypotheses and it is the m<span>ost commonly used explicit representation of atoms. </span>
Answer: a) pH of a 0.1 M vinegar solution is 2.9
b) It is an acid as pH is less than 7
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.
As vinegar is a weak acid, its dissociation is represented as;
cM 0 0
So dissociation constant will be:
Give c= 0.1 M
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.1\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.1%5Ctimes%20%5Calpha)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[1.3\times 10^{-3}]=2.9](https://tex.z-dn.net/?f=pH%3D-log%5B1.3%5Ctimes%2010%5E%7B-3%7D%5D%3D2.9)
Thus pH of a 0.1 M vinegar solution is 2.9
As pH is less than 7, it is an acid.
Answer:
0.670 mol
Explanation:
Step 1: Write the balanced reaction for the decomposition of alumina
2 Al₂O₃ ⇒ 4 Al + 3 O₂
Step 2: Calculate the moles corresponding to 34.2 g of Al₂O₃
The molar mass of Al₂O₃ is 101.96 g/mol.
34.2 g × 1 mol/101.96 g = 0.335 mol
Step 3: Calculate the moles of Al produced from 0.335 moles of Al₂O₃
The molar ratio of Al₂O₃ to Al is 2:4.
0.335 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃) = 0.670 mol Al
Answer:
- Add AgNO₃ solution to both unlabeled flasks: based on solubility rules, you can predict that when you add AgNO₃ to the NaCl solution, you will obtain AgCl precipitate, while no precipitate will be formed from the NaClO₃ solution.
Explanation:
<u>1. Adding AgNO₃ to NaCl solution:</u>
- AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
<u>2. Adding AgNO₃ to NaClO₃ solution</u>
- AgNO₃ (aq) + NaClO₃ (aq) → AgClO₃ (aq) + NaNO₃ (aq)
<u />
<u>3. Relevant solubility rules for the problem.</u>
- Although most salts containing Cl⁻ are soluble, AgCl is a remarkable exception and is insoluble.
- All chlorates are soluble, so AgClO₃ is soluble.
- Salts containing nitrate ion (NO₃⁻) are generally soluble and NaNO₃ is not an exception to this rule. In fact, NaNO₃ is very well known to be soluble.
Hence, when you add AgNO₃ to the NaCl solution the AgCl formed will precipitate, and when you add the same salt (AgNO₃) to the AgClO₃ solution both formed salts AgClO₃ and NaNO₃ are soluble.
Then, the precipiate will permit to conclude which flask contains AgCl.
