Answer:
1.67mol/L
Explanation:
Data obtained from the question include:
Mole of solute (K2CO3) = 5.51 moles
Volume of solution = 3.30 L
Molarity =?
Molarity is simply the mole of solute per unit litre of the solution. It can be expressed mathematically as:
Molarity = mole of solute /Volume of solution
Molarity = 5.51 mol/3.30 L
Molarity = 1.67mol/L
Therefore, the molarity of K2CO3 is 1.67mol/L
Explanation:
Scientist use trees a whole lot to look at climate of the past by examining tree rings.
These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.
Tree rings can be used to decipher the age of a tree.
- These three rings can be used to interpret climatic patterns.
- During a wet climate, the tree rings are more robust and bigger.
- In a dry climate, the rings are thinner.
- These alternating patterns can be used to decipher the climatic signatures in a tree.
- Sometimes, it is possible to evaluate some certain isotopes that are useful in climatic studies.
learn more:
Climate change brainly.com/question/7824762
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Answer:
highest first ionization energy - Br
smallest atomic size - Br
most metallic character - Na
Explanation:
Ionization energy of Br is higher than P because higher zeff value
Sodium is more metallic than calcium because it is able to loose electron more readily as compared to calcium because of higher electro positivity.
Atomic radius of Br is the smallest as its atomic radius is 114, P (115), Ca (197), Na (186)
(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>.
