Answer:
HCO₂/H₂O is not the acid-base conjugate pair.
Explanation:
<em>Acid and conjugate base pairs differ by an H+ ion.</em>
Neither HCO₂ nor H₂O has lost or gained protons.
The conjugate acid of H₂O is H₃O⁺
The conjugate base of HCO₃⁻ is CO₃²⁻
[A conjugate acid has one more H⁺ than its base]
Answer: 4.41 atm
Explanation:
Given that,
Original pressure of oxygen gas (P1) = 5.00 atm
Original temperature of oxygen gas (T1) = 25°C
[Convert 25°C to Kelvin by adding 273
25°C + 273 = 298K
New pressure of oxygen gas (P2) = ?
New temperature of oxygen gas (T2) = -10°C
[Convert -10°C to Kelvin by adding 273
-10°C + 273 = 263K
Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law
P1/T1 = P2/T2
5.00 atm /298K = P2/263K
To get the value of P2, cross multiply
5.00 atm x 263K = 298K x V2
1315 atm•K = 298K•V2
V2 = 1315 atm•K / 298K
V2 = 4.41 atm
Thus, the new pressure inside the canister is 4.41 atmosphere
Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
![\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c](https://tex.z-dn.net/?f=%5Cln%20P%20%3D%20%5Cdfrac%7B-%5CDelta%7BH_%7Bvap%7D%7D%7D%7BRT%7D%20%2B%20c%20)
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>
<u></u>
Answer:
![\large \boxed{\text{77.4 mL}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B77.4%20mL%7D%7D)
Explanation:
Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O
V/mL: 249
c/mol·L⁻¹: 0.0443 0.285
1. Calculate the moles of Ba(OH)₂
![\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Ba%28OH%29%24_%7B2%7D%24%7D%20%3D%20%5Ctext%7B0.249%20L%20Ba%28OH%29%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B0.0443%20mol%20Ba%28OH%29%7D_%7B2%7D%7D%7B%5Ctext%7B1%20L%20Ba%28OH%29%24_%7B2%7D%24%7D%7D%20%3D%20%5Ctext%7B0.011%2003%20mol%20Ba%28OH%29%7D_%7B2%7D)
2. Calculate the moles of HCl
The molar ratio is 2 mol HCl:1 mol Ba(OH)₂
![\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20HCl%7D%20%3D%20%5Ctext%7B0.011%2003%20mol%20Ba%28OH%29%7D_%7B2%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mol%20HCl%7D%7D%7B%5Ctext%7B1%20%20mol%20Ba%28OH%29%7D_%7B2%7D%7D%20%3D%20%5Ctext%7B0.022%2006%20mol%20HCl%7D)
3. Calculate the volume of HCl
![V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}](https://tex.z-dn.net/?f=V_%7B%5Ctext%7BHCl%7D%7D%20%3D%20%5Ctext%7B0.022%2006%20mol%20HCl%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20L%20HCl%7D%7D%7B%5Ctext%7B0.285%20mol%20HCl%7D%7D%20%3D%20%5Ctext%7B0.0774%20L%20HCl%7D%20%3D%20%5Ctextbf%7B77.4%20mL%20HCl%7D%5C%5C%5C%5C%5Ctext%7BYou%20must%20add%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B77.4%20mL%7D%7D%24%20of%20HCl.%7D)
Answer:
The next high tide will happen at 6 pm.
Explanation:
The rotation of the Earth and the translation movement of the Moon (ie the Moon's movement around the Earth), in addition to the gravitational force, contribute to the formation of the tides. The high tide will occur on the face of the Earth which is facing / facing the Moon; the low tide will occur on the faces that are forming approximately an angle of 90 ° with respect to the Moon.
There are usually two periods of high tide and two periods of low tide over a day. The interval between high tide and low tide is approximately 6 hours.
Besides the gravitational force exerted by the Moon, the tide is also influenced by the gravitational force that the Sun exerts on the Earth. The influence of the Sun is smaller due to its distance, but it can also be noticed depending on the phase of the Moon.