Answer:
The answer to your question is 3.69 x 10²¹ atoms of gold
Explanation:
Data
density = 19.32 g/cm³
dimensions = 2.5 cm x 2.5 cm x 0.1 mm
number of atoms of gold = x
Process
1.- Find the volume of the foil
Volume = length x height x width
Volume = 2.5 x 2.5 x 0.01
Volume = 0.0625 cm³
2.- Calculate the mass of the foil
Density = mass/volume
mass = density x volume
mass = 19.32 x 0.0625
mass = 1.208 g
3.- Calculate the number of atoms
Atomic number of Gold = 197 g
197 g -------------------- 6.023 x 10²³ atoms
1.208 g --------------- x
x = (1.208 x 6.023 x 10²³)/197
x = 3.69 x 10²¹ atoms of gold
The vacuoles are the answer
Do you have a picture to go with it?
Answer:
1.14 atm and 1.139 mol
Explanation:
The <em>total pressure</em> of the container is equal to the <u>sum of the partial pressure of the three gasses</u>:
- P = Poxygen + Pnitrogen + Pcarbon dioxide
- 2.50 atm = 0.52 + 0.84 + Pcarbon dioxide
Now we <u>solve for the pressure of carbon dioxide</u>:
- Pcarbon dioxide = 1.14 atm
To c<u>alculate the number of CO₂ moles </u>we use <em>PV=nRT</em>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 32 °C ⇒ 32 + 273.16 = 305.16 K
1.14 atm * 25.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305.16 K
3024.75 Joules needed to warm iron