Answer:
See figure 1
Explanation:
In this question, we have to start with the <u>protonation of the double bond</u>. In carvone we have two double bonds, so, we have to decide first which one would be protonated.
The problem states that the <u>terminal alkene</u> is the one that would is protonated. Therefore, we have to do the <u>protonation</u> in the double bond at the bottom to produce the <u>carbocation number 1</u>. Then, a hydride shift takes place to produce the <u>carbocation number 2</u>. A continuation, an <u>elimination reaction</u> takes place to produce the <u>conjugated diene</u>. Then the diene is protonated at the <u>carbonyl group</u> and with an elimination reaction of an hydrogen in the <u>alpha carbon</u> we can obtain <u>carvacol. </u>
The answer 1) Oxygen. And another smallest radius is fluorine, because it is the first in the periodic table.
Answer:
In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as
c=4.18Jg∘C
Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.
Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of 1 g of that substance by 1∘C.
In water's case, you need to provide 4.18 J of heat per gram of water to increase its temperature by 1∘C.
What if you wanted to increase the temperature of 1 g of water by 2∘C ?
This will account for increasing the temperature of the first gram of the sample by n∘C, of the the second gramby n∘C, of the third gram by n∘C, and so on until you reach m grams of water.
And there you have it. The equation that describes all this will thus be
q=m⋅c⋅ΔT , where
q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature
In your case, you will have
q=100.0g⋅4.18Jg∘C⋅(50.0−25.0)∘C
q=10,450 J
Answer:
5.37 × 10⁻⁴ mol/L
Explanation:
<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.00154 mol/L
- Initial volume (V₁): 230. mL
- Final concentration (C₂): ?
- Final volume (V₂): 660. mL
Step 2: Calculate the concentration of the final solution
We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L
