We observe that heat capacity of salted water we will find that it is less than pure water. We now that it takes less energy to increase the temperature of the salt water 1°C than pure water. Which means that the salted water heats up faster and eventually reaches to its boiling point first.
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<h3>
Answer:</h3>
12 years
<h3>
Explanation:</h3>
We are given;
Half life of hydrogen-3 is 12 years
Initial mass of Hydrogen-3 is 20 grams
Final mass will be 10 g because we are told half of the sample will decay.
To find the time taken for the decay we need to know what half life is;
- Half life is the time taken for a radioactive isotope to decay to half its original amount.
Remaining mass = Original mass × 0.5^n
n = number of half lives
therefore;
10 g = 20 g × 0.5^n
0.5 = 0.5^n
n = log 0.5 ÷ log 0.5
= 1
But, 1 half life is 12 years
Therefore, the time taken is 12 years
The answer is asparagine interactions between aspartate and arginine side chains at neutral pH
<h3>
What side chain does arginine have?</h3>
- The amino acid arginine has the formula (H2N)(HN)CN(H)(CH2)3CH(NH2)CO2H.
- A guanidino group is attached to a typical amino acid framework in the molecule.
- The carboxylic acid is deprotonated (CO2) and both the amino and guanidino groups are protonated, resulting in a cation at physiological pH.
- The guanidine group in arginine is the precursor for nitric oxide production. it is a white, water-soluble solid, like all amino acids.
learn more about arginine refer
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Answer:
4-chloro-4-methyl-cyclohexene.
Explanation:
Hello,
On the attached picture you will find the chemical reaction forming the required product, 4-chloro-4-methyl-cyclohexene. In this case, according to the Markovnicov’s rule, it is more likely for the chlorine to be substituted at the carbon containing the methyl radical in addition to the hydrogen to the next carbon to break the double bond and yield the presented product.
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Answer:
pH after the addition of 10 ml NaOH = 4.81
pH after the addition of 20.1 ml NaOH = 8.76
pH after the addition of 25 ml NaOH = 8.78
Explanation:
(1)
Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles = 2 x 10⁻³ moles,
Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles
CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O
Initial conc. 2 x 10⁻³ 1 x 10⁻³ 0
Equilibrium 1 x 10⁻³ 0 1 x 10⁻³
Final volume = 20 + 10 = 30 ml = 0.03 lit
So final concentration of Acid = 
Final concentration of conjugate base [CH₃CH₂CH₂COONa]
Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .
Using Henderson Hasselbalch equation to find the pH
![pH=pK_{a}+log\frac{[conjugate base]}{[acid]} \\\\=-log(1.54X10^{-5} )+log\frac{0.03}{0.03} \\\\=4.81](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%2Blog%5Cfrac%7B%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D%20%20%5C%5C%5C%5C%3D-log%281.54X10%5E%7B-5%7D%20%29%2Blog%5Cfrac%7B0.03%7D%7B0.03%7D%20%5C%5C%5C%5C%3D4.81)
