The answer is B. Molecules move more quickly as temperature increases.
When Allmond molecular motion stops, that is considered absolute zero. That does not mean that it cannot get colder, disapproving A.
C is just wrong.
D says when molecular motion stops the temperature STARTS to decrease, it was decreasing before it got there.
The chemical formula for chlorophyll is C55H72O5N4<span>Mg. Only 1 of the 137 atoms comes from magnesium.
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Answer : Option 1) The true statement is each carbon-oxygen bond is somewhere between a single and double bond and the actual structure of format is an average of the two resonance forms.
Explanation : The actual structure of formate is found to be a resonance hybrid of the two resonating forms. The actual structure for formate do not switches back and forth between two resonance forms.
The O atom in the formate molecule with one bond and three lone pairs, in the resonance form left with reference to the attached image, gets changed into O atom with two bonds and two lone pairs.
Again, the O atom with two bonds and two lone pairs on the resonance form left, changed into O atom with one bond and three lone pairs. It concludes that each carbon-oxygen bond is neither a single bond nor a double bond; each carbon-oxygen bond is somewhere between a single and double bond.
Also, it is seen that each oxygen atom does not have neither a double bond nor a single bond 50% of the time.
Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole


Moles = 0.246

Volume = 280 mL = 0.280 L

Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g

Moles = 1.98
Volume = 2.6 L


Molarity = 0.762 M
Molarity = 0.76 M
Answer:
3. 3.45×10¯¹⁸ J.
4. 1.25×10¹⁵ Hz.
Explanation:
3. Determination of the energy of the photon.
Frequency (v) = 5.2×10¹⁵ Hz
Planck's constant (h) = 6.626×10¯³⁴ Js
Energy (E) =?
The energy of the photon can be obtained by using the following formula:
E = hv
E = 6.626×10¯³⁴ × 5.2×10¹⁵
E = 3.45×10¯¹⁸ J
Thus, the energy of the photon is 3.45×10¯¹⁸ J
4. Determination of the frequency of the radiation.
Wavelength (λ) = 2.4×10¯⁵ cm
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
Next, we shall convert 2.4×10¯⁵ cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
2.4×10¯⁵ cm = 2.4×10¯⁵ cm × 1 m /100 cm
2.4×10¯⁵ cm = 2.4×10¯⁷ m
Thus, 2.4×10¯⁵ cm is equivalent to 2.4×10¯⁷ m
Finally, we shall determine the frequency of the radiation by using the following formula as illustrated below:
Wavelength (λ) = 2.4×10¯⁷ m
Velocity (c) = 3×10⁸ m/s
Frequency (v) =?
v = c / λ
v = 3×10⁸ / 2.4×10¯⁷
v = 1.25×10¹⁵ Hz
Thus, the frequency of the radiation is 1.25×10¹⁵ Hz.