Use the formula E=mc^2
energy given=<span>8.1 x 10^16 joules
</span>speed of <span>light = 3.00 × 10^8 m/s
</span>
plug in the values we'll get mass=<span>9.0 x 10-1 kg</span>
Answer:
The answer to your question is 1.36 x 10²³ atoms
Explanation:
Data
number of atoms = ?
mass of the sample = 34.2 g
Molecule = Cl₂O₅
Process
1.- Calculate the molar mass of Cl₂O₅
Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g
2.- Calculate the atoms of Cl₂O₅
151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms
34.2 g of Cl₂O₅ ------------ x
x = (34.2 x 6.023 x 10²³) / 151
x = 1.36 x 10²³ atoms
Answer:
2.8 * 10^(-6) / 1.4 * 10^(-2)=
2* 10^(-8)
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
