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3 years ago

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You and your roommates have already decided to spend a week volunteering in the relief effort over the winter break, but in orde

r to convince your roommates to write to Congress to demand action on climate change, you want to give them a sense of how much water fell. To do this, you calculate how many gallons of water would be in your dorm room if 52 inches of rain landed in this space. Assume that your dorm room is 228 ft^2 and know that 1 gallon = 231 in^3 and that 1 foot = 12 inches.

2 answers:

Vlada [557]3 years ago

5 0

Answer:

I dont know the answer for that question it's hard question isn't it

marta [7]3 years ago

4 0

Answer:

Gallons of water in the dorm room =7391 gallons

Explanation:

<u>Given:</u>

Amount of rainfall = 52 inches

Area of the room = 228 ft^2

<u>To determine:</u>

The gallons of water in the room

<u>Calculation:</u>

Convert area of room from ft^2 to inches^2

1 foot = 12 inches

Therefore, $1 ft^{2} = (12\ in)^{2} = 144\ in^{2}$

Area of room is:

$\frac{228ft^{2}*144in^{2}}{1ft^{2} } =32832in^{2}$

Volume of water in the room is:

$=Area\ of\ room * Amount\ of\ rainfall\\\\= 32832 in^{2} * 52\ in = 1707264\ in^{3}$

Convert from in^3 to gallons:

1 gallon = 231 in^3

Therefore:

$\frac{1707264in^{3} * 1 gallon}{231in^{3}} =7391 gallons$

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A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

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At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ 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,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

$\frac{V}{T}=k$

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 $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

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:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

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 .

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1) If the pressure on 2.6 liters of a gas at 860 Torr is increased to 1000 Torr,

LenKa [72]

2.24 liters is the volume of the gas if pressure is increased to 1000 Torr.

Explanation:

Data given:

Initial volume of the gas V1 = 2.6 liters

Initial pressure of the gas P1 = 860 Torr 1.13 atm

final pressure on the gas P2 = 1000 Torr 1.315 atm

final volume of the gas after pressure change V2 =?

From the data given above, the law used is :

Boyles Law equation:

P1V1 = P2V2

V2 = P1V1/P2

= 1.13 X 2.6/ 1.31

= 2.24 Liters

If the pressure is increased to 1000 Torr or 1.315 atm the volume changes to 2.24 liters. Initially the volume was 2.6 litres and the pressure was 860 torr.

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