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Alla [95]
3 years ago
10

Federico has two samples of pure water—sample X and sample Y. Sample X has a volume of 1 L, and sample Y has a volume of 10 L. H

ow do the boiling points of these two samples compare
Chemistry
1 answer:
Lostsunrise [7]3 years ago
8 0

Answer:

The boiling point of sample X and sample Y are exactly the same.

Explanation:

The difference between sample X and sample Y is that they occupy different volumes. However, they both contain pure water. Remember that pure water has uniform composition irrespective of its volume.

Volume does not affect the boiling point as long as the volume is small enough not to give rise to significant pressure changes in the liquid.

The boiling point of a liquid is the temperature at which the pressure exerted by the surroundings upon a liquid is equaled by the pressure exerted by the vapour of the liquid; under this condition, addition of heat results in the transformation of the liquid into its vapour without raising the temperature.

It can be clearly seen from the above that the volume of a solution of pure water does not affect its boiling point hence sample X and sample Y will have the same boiling point.

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1 year ago
PLEASE HELP ASAP!!!
alexdok [17]

Answer:

Answers are bellow.

Explanation:

The element with electron configuration 1s22s22p63s1 belong I A group in the periodic table and it is sodium because it loses one electron.

We have periodic table in attachments.

Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table, because it has a single electron in its outer shell, which it readily donates, creating a positively charged ion—the Na+ cation. Its only stable isotope is 23Na.

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8 0
2 years ago
Read 2 more answers
State Henry's law. Give an example of Henry's law.
ruslelena [56]

Answer:

Found this off of google, "Henry's law comes into play every time a bottle of Pepsi (or any other carbonated drink) is opened. The gas above the unopened carbonated drink is usually pure carbon dioxide, kept at a pressure which is slightly above the standard atmospheric pressure."

4 0
2 years ago
Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)
tangare [24]

Answer:

\boxed{\text{23.4 mmHg}}

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

K_{\text{p}} = p_{\text{H2O}}

\text{The relationship between $\Delta G^{\circ}$ and $K_{\text{ p}}$ is}\\\Delta G^{\circ} = -RT \ln K_{\text{p}}

Data:  

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

\begin{array}{rcl}8600 & = & -8.314 \times 298.15 \ln K \\8600 & = & -2478.8 \ln K\\-3.47 & = & \ln K\\K&=&e^{-3.47}\\& = & 0.0311\end{array}

Standard pressure is 1 bar.

p_{\text{H2O}} = \text{0.0311 bar} \times \dfrac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\\\\\text{The vapour pressure of water at $25 ^{\circ}\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}

4 0
2 years ago
Citric acid is often used as an acidity regulator in hot water canning of tomatoes. At 25°C it has Kat 7.4x10- and A.Hº = +4.1 k
gavmur [86]

Answer:

The rate constant at T = 100 C is 1.0*10⁻³

Explanation:

The Arrhenius equation relates two rate constants K1 and K2 measured at temperatures T1 and T2 as shown below:

ln\frac{K_{2}}{K_{1}}=\frac{\Delta H^{0}rxn}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})

here, ΔHrxn = standard enthalpy change of the reaction

R = gas constant

From the given information:

K1 = 7.4*10^-4

T1 = 25 C = 25+273 = 298 K

T2 = 100 C = 100+273 = 373K

ΔH°=4.1kJ/mol

ln\frac{K_{2}}{7.4*10^{-4}}=\frac{4.1 kJ/mol}{0.08314kJ/mol.K}(\frac{1}{298}-\frac{1}{373})K

K2 = 1.03*10⁻³

6 0
3 years ago
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