Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
C. a change of state. It can be a physical or a chemical state of change.
Answer:
3.861x10⁻⁹ mol Pb⁺²
Explanation:
We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.
We<u> calculate the mass of lead ion in 100 mL of water</u>:
- 100.0 mL ⇒ 100.0 / 1000 = 0.100 L
- 0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²
Now we <u>convert mass of lead to moles</u>, using its molar mass:
- 8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g
- 8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²
22 Newton’s to the right but I’m not to sure make sure to check other answers!
Probably sodium Na because it has a similar atomic mass and number
