Answer:
Sprinkling of powder on the carom board <u>reduces</u> friction.
Answer: 1.80g
Explanation:
Molar Mass of AlCl3 = 27 + (3x35.5)
= 27 + 106.5 = 133.5g/mol
Number of mole of AlCl3 = 0.0135mol
Mass = 0.0135 x 133.5= 1.80g
The balloon will reach its maximum volume and it will burst.
Given:
- A weather balloon at sea level, with gas at 65.0 L volume, 745 Torr pressure, and 25C temperature.
- When the balloon was taken to an altitude at which temperature was 25C and pressure was 0.066atm its volume expanded.
- The maximum volume of the weather balloon is 835 L.
To find:
Whether the weather balloon will reach its maximum volume or not.
Solution:
The pressure of the gas in the weather balloon at sea level = 

The volume of the weather balloon at sea level = 
The temperature of the gas in the weather balloon at sea level:

The balloon rises to an altitude.
The pressure of the gas in the weather balloon at the given altitude:

The volume of the weather balloon at the given altitude = 
The temperature of the gas in the weather balloon at the given altitude:

Using the Combined gas law:

The maximum volume of the weather balloon= V = 835 L

The volume of the weather balloon at a given altitude is greater than its maximum volume which means the balloon will reach its maximum volume and it will burst.
Learn more about the combined gas law:
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Heterogeneous mixture. cause you can separate the noodles from soup
Answer:
mass of oxygen gas in Kg = 15.0Kg
Explanation:
Volume of air in the room = 4.0m*5.0m*2.5m = 50m³
volume of oxygen in the room = 21/100 * 50m³ = 10.5m³
using the ideal gas equation; PV=nRT
number of moles of oxygen gas, n = PV/RT
At STP, P = 1atm, V = 10.5m³ = (10.5*1000)dm³ = 10500dm³, R = 0.082 atmdm³K⁻¹mol⁻¹, T = 273K
n = 1 * 10500/ (273 *0.082)
n = 469.04 moles
mass of oxygen gas in Kg = (no of moles * molar mass)/1000
molar mass of oxygen gas = 32g
mass of oxygen gas in Kg = (469.04 * 32)/1000
mass of oxygen gas in Kg = 15.0Kg