Casts! Molds! Trace Fossils as well
Answer:
836.8J
Explanation:
Heat energy = ?
Mass = 100g
Initial temperature (T1) = 21°C
Final temperature (T2) = 23°C
Specific heat capacity of water (c) = 4.184J/g°C
To solve this question, we'll need to use the formula for calculating heat energy of a substance.
Q = mc∇t
Q = heat energy of substance
M = mass of substance
C = specific heat capacity of substance
∇t = change in temperature = T2 - T1
Q = 100 × 4.184 × (23 -21)
Q = 418.4 × 2
Q = 836.8J
The heat energy required to raise 100g of water from 21°C to 23°C is 836.8J
Answer:
The amount of heat required to raise the temperature of the sample from 298 to 385 Kelvin, is 16.6 kJ
Explanation:
<u>Step 1: </u>Given data
A 79.0 g sample of ethanol raises from 298 K to 385 K
The specific heat of ethanol is 2.42J/g°C
<u>Step 2:</u> Calculate the heat transfer
Q = m*Cp*ΔT
with m = the mass of the ethanol sample (in grams)
⇒ mass = 79 grams
with Cp = the specific heat capacity of ethanol (in J/g°C)
⇒ Cp = 2.42 J/g°C
with ΔT = the change of temperature (T2-T1)
⇒ ΔT = 385 K - 298K = 112 °C - 25 °C = 87
Q = 79 grams * 2.42 J/g°C * 87 = 16632.66 j = 16.6 kJ
The amount of heat required to raise the temperature of the sample from 298 to 385 Kelvin, is 16.6 kJ
An exothermic reaction has occurred. You can remember that EXothermic means that energy is EXiting the system. If energy enters the system, it’s endothermic.
Answer:
The compound elucidated from the spectral data is <u>4-methyl penta-2-none</u>
Explanation:
- <u>1700 cm-1 from IR data</u> suspects aldehyde/ketone or carboxylic acid. However,since the peak is not a stretched vibration, it implies an aldehyde or ketone present.
- <u>3 ppm H NMR</u> confirms O-CH3 bond
- <u>24.4 13C NMR</u> confirms CH3-R bond
- 26.4 13C NMR confirms H3C-(R)C(H)-R
- 44.2 13C NMR Confirms C=
- 2126 13C NMR confirms aldehyde C=O bond
<u>The deduced structure is 4-methyl penta-2-one (see attached) </u>given multiple CH3 atoms.