Answer is: <span>C. propanal.
1-propanol is primary alcohol. With weak oxidizing agent primary alcohol gives aldehyde and with strong oxidizing agent primary alcohol gives carbonic acid.
</span>Oxidation of 1-propanol with <span>Na</span>₂<span>Cr</span>₂<span>O</span>₇<span> and sulfuric acid</span><span> gives propanal and o</span><span>xidation with </span>chromic acid<span> gives </span>propionic acid<span>.</span>
Carbon -13 has 7 neutrons and carbon -12 has six neutrons. Carbon -12 is the most common isotope of Carbon. Carbon -14 is radioactive and vary rare. The symbols for the isotopes of Carbon atoms shown here indicate they each have six protons but mass numbers of 14, 13, and 12. Hope this helps. :)
Increasing order of strength needed to break bonds:
temporary dipole induced dipole interactions
Permanent dipole induced dipole interactions
Hydrogen bonding
The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months
To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Amount remaining (N) = 5%
Original amount (N₀) = 100%
<h3>Number of half-lives (n) =?</h3>
N₀ × 2ⁿ = N
5 × 2ⁿ = 100
2ⁿ = 100/5
2ⁿ = 20
Take the log of both side
Log 2ⁿ = log 20
nlog 2 = log 20
Divide both side by log 2
n = log 20 / log 2
<h3>n = 4.32</h3>
Thus, 4.32 half-lives gas elapsed.
Finally, we shall determine the half-life of the element. This can be obtained as follow.
Number of half-lives (n) = 4.32
Time (t) = 500 years
<h3>Half-life (t½) =? </h3>
t½ = t / n
t½ = 500 / 4.32
t½ = 115.74 years
Multiply by 12 to express in months
t½ = 115.74 × 12
<h3>t½ ≈ 1389 months </h3>
Therefore, the half-life of the radioactive element in months is approximately 1389 months
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