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MakcuM [25]
3 years ago
10

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

give brainiest.
Chemistry
1 answer:
Ghella [55]3 years ago
7 0

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>

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Answer:

Si las condiciones para que el magma permanezca líquido no perduran, el magma se enfriará y solidificará en una roca ígnea. Una roca que se enfría en el interior de la Tierra se denomina intrusiva o plutónica y su enfriamiento será muy lento, produciendo una estructura cristalina de granos grueso.

Explanation:

7 0
3 years ago
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
LuckyWell [14K]
To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

        = -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831 

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m  

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

       = -40

     
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3 years ago
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A sample has a mass of 35.4g and a volume of 36.82mL. What is the density of the sample
vredina [299]

The density of the sample is 0.96 g/mL.


Density is found by dividing the mass of an object by its volume.


D=m÷v

D=35.4 g÷36.82 mL

D=0.96 g/mL

6 0
3 years ago
Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base
Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0
3 years ago
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