Answer: 70.0°C
Explanation:
Quantity of heat = Mass * Specific heat * Change in temperature
Quantity of heat = 104.6 KJ
Mass = 500.0 g
Specific heat of water is 4.18 J/g°C
Change in temperature assuming final temperature is x = x - 20
Units should be in grams and joules:
104,600 = 500 * 4.18 * (x - 20)
104,600 = 2,090 * (x - 20)
x - 20 = 104,600/2,090
x = 104,600/2,090 + 20
x = 69.8
= 70.0°C
Answer:
The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g
Explanation:
Step 1: Given data
The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)
The sample hasa volume of 25.0 mL
Step 2: Calculating mass of the sample
The density is the mass per amount of volume
0.469g/cm³ = 0.469g/ml
The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g
The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g
1 gram of sugar because super molecules are bigger then the ions of dissolved salt
