Use the universal gas formula
PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)
Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)
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Answer:
130
Explanation:
This is because that 3atm of N2O4 is used up for the 6atm of NO2, so 1 atm N2O4 is left. Resulting in In(1/4).
Answer:
3.0 × 10²⁰ molecules
Explanation:
Given data:
Mass of ethanol = 2.3 × 10⁻²°³ g
Number of molecules = ?
Solution:
Number of moles of ethanol:
Number of moles = mass/ molar mass
Number of moles = 2.3 × 10⁻²°³ g / 46.07 g/mol
Number of moles = 0.05 × 10⁻²°³ mol
Number of molecules:
One mole = 6.022 × 10²³ molecules
0.05 × 10⁻²°³ mol × 6.022 × 10²³ molecules / 1 mol
0.30 × 10²⁰°⁷ molecules
3.0 × 10¹⁹°⁷ molecules which is almost equal to 3.0 × 10²⁰ molecules.
590 mL = 590 cm³= 0,59 dm³
C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
_____________________________
M KNO₃ = 39g+14g+16g×3 = 101 g/mol
1 mole -------- 101g
0,649 --------- X
X = 101×0,649
X = 65,549g KNO₃
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