Answer:
<em><u>= - 0.38 eV</u></em>
Explanation:
Using Bohr's equation for the energy of an electron in the nth orbital,
E = -13.6 
Where E = energy level in electron volt (eV)
Z = atomic number of atom
n = principal state
Given that n = 6
⇒ E = -13.6 × 
<em><u>= - 0.38 eV</u></em>
<em><u></u></em>
<em>Hope this was helpful.</em>
<em><u></u></em>
Answer is: volume of KBr is 357 mL.
c(KBr) = 0,716 M = 0,716 mol/L.
m(KBr) = 30,5 g.
n(KBr) = m(KBr) ÷ M(KBr).
n(KBr) = 30,5 g ÷ 119 g/mol.
n(KBr) = 0,256 mol.
V(KBr) = n(KBr) ÷ c(KBr).
V(KBr) = 0,256 mol ÷ 0,716 mol/L.
V(KBr) = 0,357 L · 1000 mL/L = 357 mL.
Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature
a. It is likely that more rock candy will be formed in batch A.
b. It is likely that less rock candy will be formed in batch A.
c. It is likely that no rock candy will be formed in either batch.
d. I need more information to predict which batch is more likely to form rock candy.
Answer: Option A
Explanation:
More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.
Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.
The sugar will be dissolved in water until the time all the space is filled sugar molecules.
Hence, the correct answer is Option A.
Answer:
Speed of Sound
Explanation:
Speed of sound, fluid phases
m/s
notes
WEL 206
use 323 27 °C
CRC 323 27 °C
WEL 319
Electrons in sigma <span>bonds remain localized between two atoms. Sigma </span><span>bond results from the formation of </span><span>a molecular orbital </span><span>by the end to </span><span>end overlap of atomic </span>orbitals. Electrons<span> in pi</span> bonds can become delocalized between more than two atoms. Pi bonds result from the formation of molecular orbital by side to side overlap of atomic orbitals.
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