Answer:
Ethyl Acetate with a chemical configuration of : CH3CO2CH2CH3
Explanation:
Ethyl Acetate has a density of 902 kg/m³, molar mass of 88.11 g/mol and posses a boiling point of 77.1 °C.
It is to be understood that ethyl acetate is the ester of ethanol and it is mostly produced in mass production and most especially used in the production of domestic materials especially materials like glue, efostics and dissolving agent. This is a highly toxic and flammable substance. However colorless and possess a sweet smell, it can be highly poisonous when ingested.
B obviously not a or c so if it aint b its D
From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.
From the information we have;
Volume of the damp air = 1 L
Pressure of the damp air = 741.0 torr or 0.975 atm
Temperature of the gas = 20 oC + 273 = 293 K
R = 0.082 atm LK-1mol-1
Number of moles = ?
n =PV/RT
n = 0.975 × 1/0.082 × 293
n = 0.041 moles
Volume of water vapor = 1 L
Temperature of water = -10 oC + 273 = 263 K
Pressure of the gas = 607.1 torr or 0.799 atm
R = 0.082 atm LK-1mol-1
n= PV/RT
n = 0.799 × 1/ 0.082 × 263
n = 0.037 moles
Number of moles of water = 0.041 moles - 0.037 moles = 0.004 moles
If 1 mole = 6.02 × 10^23 molecules
0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole
= 2.41 × 10^21 molecules
Learn more: brainly.com/question/2510654
Answer:
(A) 15.0 °C
Explanation:
The water in beaker A gains heat because its initial temperature (10 °C) is less than the initial temperature of the water in beaker B (20 °C) which loses heat.
Let T3 be the final temperature
Heat gained by beaker A = heat loss by beaker B
mc(T3 - T1) = mc(T2 - T3)
The mass and specific heat of water in both beakers are the same. Therefore, (T3 - T1) = (T2 - T3)
T1 is initial temperature of beaker A = 10 °C
T2 is initial temperature of beaker B = 20 °C
T3 - 10 = 20 - T3
T3 + T3 = 20 + 10
2T3 = 30
T3 = 30/2 = 15 °C
A is a model of a decomposition reaction
