4 Movement of less dense material
3 Heating of cooler material
1 cooling of warmer material
2 movement of denser material
Answer:
0.1593 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.
P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.
<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>
Answer:
8.70 liters
Explanation:
First we <u>convert 36.12 g of AI₂O₃ into moles</u>, using its <em>molar mass</em>:
- 36.12 g ÷ 101.96 g/mol = 0.354 mol AI₂O₃
Then we <u>convert AI₂O₃ moles into O₂ moles</u>, using the stoichiometric coefficients of the reaction:
- 0.354 mol AI₂O₃ *
= 0.531 mol O₂
We can now use the <em>PV=nRT equation</em> to <u>calculate the volume</u>, V:
- 1.4 atm * V = 0.531 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 280.0 K
A hydrate is a substance where in it contains water and other constituent elements. To know whether if that compound was a hydrate,you should record its mass, then put it in a test tube and heat it with a Bunsen burner. If the compound is a hydrate, the water in the compound will discharge in the form of water vapor. At the next 5-10 minutes, remove it in the test tube and weigh it up again. If the mass is now fewer, that means that there was water existing that has now evaporated, and the compound was a hydrate.
Molar mass Na = 23.0 g/mol
1 mol ---- 23.0 g
n mol ---- 69 g
n = 69 / 23.0
n = 3.0 moles
1 mole -------- 6.02x10²³ molecules
3.0 moles ---- ?
3.0 * 6.02x10²³ / 1
= 1.806x10²⁴ molecules
hope this helps!
