Answer:
Kb = 0.428 m/°C
Explanation:
To solve this problem we need to use the <em>boiling-point elevation formula</em>:
- <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.
So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.
molality = mol solute / kg solvent
- Density of CS₂ = 1.26 g/cm³
- Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg
molality = 0.270 mol / 0.5166 kg = 0.5226 m
Now we <u>solve for Kb</u>:
<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
- 47.52 °C - 46.3 °C = Kb * 0.5226 m
<span>biological reactions that happen within cells while reducing the complex interactions found in a whole cell. Eukaryotic and prokaryotic cells have been used for creation of these simplified environments[1]. Subcellular fractions can be isolated by ultracentrifugation to provide molecular machinery that can be used in reactions in the absence of many of the other cellular components.
Cell-free biosystems can be prepared by mixing a number of purified enzymes and coenzymes. Cell-free biosystems are proposed as a new low-cost biomanufacturing platform compared to microbial fermentation used for thousands of years. Cell-free biosystems have several advantages suitable in industrial applications</span>
D is the correct answer
every other option contains an element
2 C2H2 + 5 02 > 4 CO2 + 2 H2O
Products - Reactants ( all units are kJ/mo1):
(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1
-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1
answer: -1255.6 kJ/mo1
Answer:
Ionization energy
Electronegativity
Explanation:
-due to its smaller ionic radius....the electron in the outter most shell tends to expierence a stronger nuclear attraction...which makes it harder to remove the electron from the sodium atom
-Rubidium has lesser ionization energy because its (i) affected by its larger ionic radius which tends to lessen the nuclear attraction ...hence making it easier to remove the electron...(ii)and also by the screening effect done by the inner shells, which also tends to lessen the nuclear attraction.
Sodium has a higher electronegativity than rubidium;
Electronegativity is the charge density of electrons in an atom...in which its high when the atomic radius is smaller...
So hence due to the sodium atomic radius being smaller...it tends to have a higher charge density than rubidium....which then gives it a higher electronegativity value
