The pressure of the gas = 40 atm
<h3>Further explanation</h3>
Given
200 ml container
P = 2 atm
final volume = 10 ml
Required
Final pressure
Solution
Boyle's Law
At a fixed temperature, the gas volume is inversely proportional to the pressure applied
Input the value :
P₂ = P₁V₁/V₂
P₂ = 2 x 200 / 10
P₂ = 40 atm
Answer:
I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong
Explanation:
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
Answer:
C: The C horizon is a subsurface horizon. It is the least weathered horizon. Also known as the saprolite, it is unconsolidated, loose parent material. The master horizons may be followed by a subscript to make further distinctions between differences within one master horizon.
Explanation:
Answer:
A. increasing the ability of technology, like microscopes, to see even smaller particles
