Answer:
- 0.99 °C ≅ - 1.0 °C.
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.
<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>
Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient: ![Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2](https://tex.z-dn.net/?f=Q_%7Br%7D%20%3D%5Cfrac%7B%5Cleft%20%5B%20PEP%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%202PG%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B0.1%20mM%7D%7B0.5%20mM%7D%20%3D%200.2)
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

![\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20kJ%2Fmol%20%2B%20%5B2.303%20%5Ctimes%20%288.314%20%5Ctimes%2010%5E%7B-3%7D%20kJ%2F%28K.mol%29%29%5Ctimes%20%28310.15%20K%29%5D%20log%20%280.2%29)
![\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20%2B%20%5B5.938%5D%20%5Ctimes%20%28-0.699%29%20%3D%201.7%20-%204.15%20%3D%20%28-2.45%20kJ%2Fmol%29)
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>
Answer:
sample A
Explanation:
the first one because of the ppm value
Answer:
The correct option is: When the amount of acid and base are equal
Explanation:
Titration is an analytic method that is used to determine the concentration of an<em> unknown solution</em>, called <em>titrand</em>.
In this method, standard solution of known concentration, called <em>titrant</em>, is taken in the burette and added drop-wise to the titrand solution in the flask, until the endpoint is reached.
In case of an acid-base titration, a <em>pH indicator</em> is used, which changes the color of the solution when the endpoint is reached.
<u>The </u><u>endpoint</u><u> indicates the </u><u>equivalence point</u><u> of an acid-base titration, where the </u><em><u>concentration of the acid and base is equal</u></em><u>. </u>
<u>Therefore, the </u><u>correct option</u><u> is: </u><u>When the amount of acid and base are equal</u>
The given statement is true .
<h3>What is Rutherford’s gold foil experiment?</h3>
- A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.
- The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.
- For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.
To learn more about Rutherford’s gold foil experiment, refer to:
brainly.com/question/4113533
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