Identify the importance of water. check all that apply
1 answer:
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At temperature, a sample of neon gas will occupy
volume.
Further Explanation:
The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”
Mathematically the law can be expressed as,
Or,
Here, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> is proportionality constant.
Given information:
The initial volume of neon gas is .
The final volume of neon gas is .
The initial temperature value is .
To calculate:
The final temperature
Given Condition:
- The pressure is constant.
- Mass of gas is fixed.
Solution:
Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:
Here,
is the initial volume of the gas.
is the final volume of the gas.
is the initial temperature of the gas.
is the final temperature of the gas.
Step 2: Rearrange equation (2) for .
…… (2)
Step 3: Convert the given temperature from degree Celsius to Kelvin.
The conversion factor to convert degree Celsius to Kelvin is,
…… (3)
Substitute for
in equation (3) to convert temperature from degree Celsius to Kelvin.
Step 4: Substitute for
,
for
and
for
in equation (2) and calculate the value of
.
Important note:
- The temperature must be in Kelvin.
- The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.
Learn More:
1. Gas laws brainly.com/question/1403211
2. Application of Charles’ law brainly.com/question/7434588
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: States of matter
Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:
Moles of glucose =
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution =
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution =
Volume of the solution taken =
Molarity of the solution after dilution =
Volume of the solution after dilution=
Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L
Moles of glucose =
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.