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bezimeni [28]
3 years ago
7

What is pomsis? can someone please help ​

Chemistry
1 answer:
Feliz [49]3 years ago
8 0
Yeah man I can help explain a little bit fits
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A 0.223 mole sample of gas is held at 33.0 C and 2.00 atm, What's the volume of the gas? R = 0.0821 L atm / mol K answer soon il
Ghella [55]

Answer:

The volume of the gas is 2.80 L.

Explanation:

An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:

P*V = n*R*T

where R is the constant of ideal gases.

In this case:

  • P= 2 atm
  • V= ?
  • n=0.223 moles
  • R= 0.0821 \frac{L*atm}{mol*K}
  • T=33 °C= 306 °K (being O°C= 273°K)

Replacing:

2 atm* V= 0.223 moles*0.0821 \frac{L*atm}{mol*K}* 306 K

Solving:

V=\frac{0.223 moles*0.0821\frac{L*atm}{mol*K} * 306 K}{2 atm} \\

V= 2.80 L

<u><em>The volume of the gas is 2.80 L.</em></u>

7 0
3 years ago
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Sun, use fusion to combine hydrogen atoms into helium atoms,
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3 years ago
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What pressure is exerted by 932.3 g of CH4 in a 0.560 L steel container at 136.2 K?
Roman55 [17]
Molar mass CH4 = 16.0 g/mol

* number of moles:

932.3 / 16 => 58.26875 moles

T = 136.2 K

V = 0.560 L

P = ?

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

P x 0.560 =  58.26875 x 0.082 x 136.2

P x 0.560 = 650.76

P = 650.76 / 0.560

P = 1162.07 atm
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3 years ago
What three factors affect the biodiversity of an ecosystem
Korolek [52]
<span>Factors that affect biodiversity in an ecosystem include area,climate,diversity of niches,and keystone species.</span><span />
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3 years ago
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A star is estimated to have a mass of 2.0 x 10 ^36kg. Assuming it to be a sphere of average radius of 7.0 x 10 ^5 km. Calculate
Montano1993 [528]

Answer:

<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>

<em>b) 8.69 x 10^7 lb/ft^3</em>

<em></em>

Explanation:

mass of the star m =  2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere = \frac{4}{3}\pi  r^{3}

==> V = \frac{4}{3}*3.142*(7.0*10^8)^{3} = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>

in lb/ft^3 =  (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>

6 0
2 years ago
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