Answer:
- <u><em>g) Neither plant should increase by 1 cm in height.</em></u>
Explanation:
See the graph for this question on the figure attached.
The growing of the <em>plant A</em> is represented by the line that goes above the other. At start, that line has a slope that rises about 0.75 cm ( height increase) in 1 day. From the day 2 and forward the slope of the line decreases. The line reaches its highest point about at day 4 and seems to start decreasing. Thus, you should predict that on the day six it <em>most likely </em>does not increase in height.
The growing of the <em>plant B</em> is represented by the line drawn below the other. As for the plant B, the growing decreases with the number of days. Between the days 4 and 5 the line is almost flat, which means that <em>most likely</em> this plant will not grow on the day six or grow less than 0.5 cm.
Thus, for both plants you can say that <em>on day six, most likley, neither should increase by 1 cm in height (</em>option g).
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
Atomic mass of an atom is defined as the sum of number of neutrons and number of protons that are present in an atom. It is represented as 'A'.
Atomic number = Number of protons + Number of neutrons
We are given:
Number of protons = 6
Number of neutrons = 6
Number of electrons = 6
Atomic mass = 6 + 6 = 12
Hence, the correct answer is Option D.
Answer:
Explanation:
The specific heats of water and steel are
Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium 
An energy balance can be written as
Replacing
Then, the temperature
Answer:
230,294.9 I think it's fine just try your past or ask teacher help
I have attached an image of the IR spectrum required to answer this question.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.
