8.03 solutions report is described below.
Explanation:
8.03 Solutions Lab Report
In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.
Pre-lab Questions:
In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.
Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).
Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.
Answer:
b- The heat capacity ratio increases but output temperature don’t change
Explanation:
The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.
Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.
On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.
Either it’s, it is released when the reaction is complete or it is changed into atoms of carbon and oxygen during the reaction
Answer:
V = 2.32 Liters
Explanation:
PV = nRT => V = nRT/P
n = 25.8g/122g/mole = 0.21 mole
R = 0.08206 L·atm/mol·K
T = 25.44°C + 273 = 298.44K
P = 2.22 atm (given in problem)
V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm
