3644000 rounded to 2 significant figures is equivalent to 3.6
The female part of the chromosomes, usually colored pink in diagrams
Answer:
We could do two 1:50 dilutions and one 1:4 dilutions.
Explanation:
Hi there!
A solution that is 1000 ug/ ml (or 1000 mg / l) is 1000 ppm.
Knowing that 1 ppm = 1000 ppb, 100 ppb is 0.1 ppm.
Then, we have to dilute the stock solution (1000 ppm / 0.1 ppm) 10000 times.
We could do two 1:50 dilutions and one 1:4 dilutions (50 · 50 · 4 = 10000). Since the first dilution is 1:50, you will use the smallest quantity of the stock solution (if we use the 10.00 ml flask):
First step (1:50 dilution):
Take 0.2 ml of the stock solution using the third dispenser (20 - 200 ul), and pour it in the 10.00 ml flask. Fill with water to the mark (concentration : 1000 ppm / 50 = 20 ppm).
Step 2 (1:50 dilution):
Take 0.2 ml of the solution made in step 1 and pour it in another 10.00 ml flask. Fill with water to the mark. Concentration 20 ppm/ 50 = 0.4 ppm)
Step 3 (1:4 dilution):
Take 2.5 ml of the solution made in step 3 (using the first dispenser 1 - 5 ml) and pour it in a 10.00 ml flask. Fill with water to the mark. Concentration 0.4 ppm / 4 = 0.1 ppm = 100 ppb.
Answer:
The reactive nucleophile is Ketone.
Explanation:
In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.
The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.
Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles
Answer:
20.11 g.
Explanation:
What is given?
c (specific heat of iron) = 0.450 J/g °C.
Q (heat energy) = 179.85 J.
ΔT (change of temperature) = |31.42 °C - 51.29 °C| = 19.87 °C.
What do we need? Mass of iron (m)
Step-by-step solution:
Let's see the formula of specific heat:
Where c is specific heat, Q is heat energy, m is mass and ΔT is the change of temperature.
We just have to solve for 'm' to find the mass of iron and replace the given data that we have, like this:
The mass of the iron would be 20.11 g.
