To find the number of moles of gas we can use the ideal gas law equation, we dont need to use the mass of gas given as we only have to find the number of moles
PV = nRT
P - pressure - 300.0 kPa
V - volume - 25.0 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 27 °C + 273 = 300 K
substituting these values in the equation
300.0 kPa x 25.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 300 K
n = 3.01 mol
number of mols of gas - 3.01 mol
Answer:
NH₃/NH₄Cl
Explanation:
We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.
![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.
Let's consider the following buffers and their pKa.
- CH₃COONa/CH3COOH (pKa = 4.74)
The optimum buffer is NH₃/NH₄Cl.
Answer:
The boiling point of water at 550 torr will be 91 °C or 364 Kelvin
Explanation:
Step 1: Data given
Pressure = 550 torr
The heat of vaporization of water is 40.7 kJ/mol.
Step 2: Calculate boiling point
⇒ We'll use the Clausius-Clapeyron equation
ln(P2/P1) = (ΔHvap/R)*(1/T1-1/T2)
ln(P2/P1) = (40.7*10^3 / 8.314)*(1/T1 - 1/T2)
⇒ with P1 = 760 torr = 1 atm
⇒ with P2 = 550 torr
⇒ with T1 = the boiling point of water at 760 torr = 373.15 Kelvin
⇒ with T2 = the boiling point of water at 550 torr = TO BE DETERMINED
ln(550/760) = 4895.4*(1/373.15 - 1/T2)
-0.3234 = 13.119 - 4895.4/T2
-13.4424= -4895.4/T2
T2 = 364.2 Kelvin = 91 °C
The boiling point of water at 550 torr will be 91 °C or 364 Kelvin
Answer:
(a) I⁻ (charge 1-)
(b) Sr²⁺ (charge 2+)
(c) K⁺ (charge 1+)
(d) N³⁻ (charge 3-)
(e) S²⁻ (charge 2-)
(f) In³⁺ (charge 3+)
Explanation:
To predict the charge on a monoatomic ion we need to consider the octet rule: atoms will gain, lose or share electrons to complete their valence shell with 8 electrons.
(a) |
I has 7 valence electrons so it gains 1 electron to form I⁻ (charge 1-).
(b) Sr
Sr has 2 valence electrons so it loses 2 electrons to form Sr²⁺ (charge 2+).
(c) K
K has 1 valence electron so it loses 1 electron to form K⁺ (charge 1+).
(d) N
N has 5 valence electrons so it gains 3 electrons to form N³⁻ (charge 3-).
(e) S
S has 6 valence electrons so it gains 2 electrons to form S²⁻ (charge 2-).
(f) In
In has 3 valence electrons so it loses 3 electrons to form In³⁺ (charge 3+).
Answer:
1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.
2. Rainwater or Damp/moist air
3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws
Explanation:
1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.
In order for galvanic corrosion to occur, three elements are required.
i. Two metals with different corrosion potentials (anode and cathode)
ii. Direct metal-to-metal electrical contact
iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.
For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.
2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.
3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.
All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.
