Answer:
Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L
Explanation:
STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.
According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .
Molar mass of Lithium Oxide = 29.8 g/mol
= 2(6.9) + 15.99 = 29.8
Mass of 1 mole = 29.8 g
1 mole occupies, Volume = 22.4 L
29.8 g occupies, V = 22.4 L
1 g occupies ,V = 
1 g occupies ,V = 0.7516 L
10 g tex]Li_{2}O[/tex] occupies ,V = 
L
V = 7.52 L
So, volume occupied by Lithium Oxide At STP is 7.52 L
Answer:
P = 995.6 atm
Explanation:
assuming ideal gas:
∴ Tst = 25°C ≅ 298 K
∴ V = 35.00 mL = 0.035 L
∴ molar mass CO2 = 44.01 g/mol
∴ mass CO2(g) = 62.76 g
⇒ mol CO2(g) = (62.76 g)*(mol/44.01 g) = 1.426 mol
∴ R = 0.082 atm.L/K.mol
⇒ P = RTn/V
⇒ P = ((0.082 atm.L/K.mol)*(298 K)*(1.426 mol)) / (0.035 L)
⇒ P = 995.6 atm
It's a gaslike state of matter consisting of positively charged ions, free electrons, and neutral particles. A hot, energetic phase of matter in which the atoms are broken apart into positive ions and negative electrons that move independently of each other.
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Answer:</h3>
51.93 L
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Explanation:</h3>
From the question we are given the following components of an ideal gas;
Number of moles = 21.5 mol
Pressure, P = 9.65 atm
Temperature, T = 10.90°C, but K= °C + 273.15
=284.05 k
We are required to calculate the volume of the ideal gas.
We are going to use the ideal gas equation which is given by;
PV = nRT, where P, V, T and n are the pressure, volume, temperature and moles of the ideal gas respectively. R is the ideal gas constant, 0.082057 L.atm/mol.K
To get the volume, we rearrange the formula to get;
V = nRT ÷ P
= (21.5 × 0.082057 × 284.05 K) ÷ 9.65 atm
= 51.93 L
Thus, the volume of the ideal gas is 51.93 L
