Answer:
Explanation:
A 12.48 g sample of an unknown metal, heated to 99.0 °C was then plunged into 50.0 mL of 25.0 °C water. The temperature of the water rose to 28.1 Go to calculating final temperature when mixing two samples of water ... Problem #1: A 610. g piece of copper tubing is heated to 95.3 °C and placed in an ... The two rings are heated to 65.4 °C and dropped into 12.4 mL of water at 22.3 °C. ... Problem #4: A 5.00 g sample of aluminum (specific heat capacity = 0.89 J g¯1
Answer:
This part require data such as Avogadro's number and the molar mass of water. But first, let's find the mass of water in the specified volume by making use of the density formula:
Density = mass/volume
1 g/mL = Mass/70 mL
Mass = 70 g
Each water contains 18 grams per mole, and each mole contains 6.022×10²³ molecules of water. Thus,
70 g * 1mole/18 g * 6.022×10²³ molecules/mole = 2.342×10²⁴ molecules of water
Explanation:
Answer:
c
Explanation:
because the ion are Mobil which mean they are free not combined and carry a charge but when they are combined/ fixed in position they can't carry a charge so therefore can't conduct electricity :)
For #4 is 298.48 hope it is correct
