Forests and prairies are examples of ecosystems on land. An ecosystem is a community of living things. Members survive by interacting with each other and with their environment. At first glance, the ocean seems like one big ecosystem.
Answer:
4.8x10⁻³ Liters are required
Explanation:
Molarity is an unit of concentration in chemistry defined as the ratio between moles of solute (In this case, silver nitrate) and liters of solution.
The 0.50M solution contains 0.50 moles of silver nitrate per liter of solution.
To provide 2.4x10⁻³ moles Silver nitrate are required:
2.4x10⁻³ moles * (1L / 0.50 moles) =
<h3>4.8x10⁻³ Liters are required</h3>
The Mesosphere, like the troposphere layer, has a decrease in temperature with altitude because of the decreases in the density of the air molecules. Thermosphere: As the altitude increases, the air temperature increases.
The balanced equation for the reaction is as follows;
Ca(OH)₂ + 2HBr --> CaBr₂ + 2H₂O
stoichiometry of Ca(OH)₂ to HBr is 1:2
number of Ca(OH)₂ moles reacted - 0.10 mol/L x 0.1000 L = 0.010 mol
Number of HBr moles added - 0.10 mol/L x 0.4000 = 0.040 mol
1 mol of Ca(OH)₂ needs 2 mol of HBr for neutralisation
therefore 0.010 mol of Ca(OH)₂ needs - 0.010 x 2 = 0.020 mol of HBr to be neutralised
but 0.040 mol of HBr has been added therefore number of moles of HBr in excess - 0.040 - 0.020 = 0.020 mol
then pH of the medium can be calculated using the excess H⁺ ions
HBr is a strong acid therefore complete ionization
[HBr] = [H⁺]
[H⁺] = 0.020 mol / (100.0 + 400.0 mL)
= 0.020 mol / 0.5 L
= 0.040 mol/L
pH = -log[H⁺]
pH = - log [0.040 M]
pH = 1.40
pH of the medium is 1.40
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
