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Trava [24]
3 years ago
13

A sample of methane gas occupies 4.50 gallons at a temperature of 80.0 F. If the pressure is held constant, what will be the vol

ume of the gas at 100. degrees C?
Chemistry
1 answer:
Marat540 [252]3 years ago
7 0

Answer:

21.19 L or 5.6 gallons

Explanation:

Data Given

initial volume V1 = 4.50 gallons

convert volume from gallons to L

                               1 gallon = 3.78541 L

                               4. 50 gallon = 3.785 L x 4.50

So,                        

initial volume V1 =  17.03 L

final Volume V2 = ?

initial Temperature T1  = 80° F

convert Temperature to Kelvin

                              T1 = ( 80° F -32) / 1.8 +273

                               T1 = 26.7° C + 273 = 299.7 K

final Temperature T2 = 100° C

convert Temperature to Kelvin

                                  T2 = 100° C + 273 = 373 K

Solution:

This problem will be solved by using charles' law equation at constant pressure.

The formula used

                      V1 / T1 = V2 / T2

As we have to find out volume, so rearrange the above equation

                      V2 = V1 x T2 / T1

Put value from the data given and conversion information in above equation

V2 =  17.03 L x 373 K / 299.7 K

V2 =  21. 19 L

So the final volume of Methane gas = 21.19 L

Or for Volume in gallons

1 L = 0.26

21.19 x 0.264 = 5.6 gallons

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When considering light moving through a diffraction grating, you should treat the light as what?
liraira [26]

Answer:

A particle

Explanation:

Modern quantum theory holds that light has both wave-like and particle-like properties. When the length scales involved are large compared to the wavelengths of light (ex., forming images with thin lenses), the

particle nature of light dominates.

5 0
3 years ago
Read 2 more answers
Why the ph of glycine increases when 0.1 M NaOH is added dropwise​
shtirl [24]

Answer:

The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.

Explanation:

The glycine is an amino acid with the following chemical formula:

NH₂CH₂COOH  

The COOH functional group is what gives the acid properties in the molecule.      

Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:

NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O

The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.  

Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.  

I hope it helps you!  

7 0
3 years ago
The ionization of pure water forms _____.
Evgesh-ka [11]

Answer: The ionization of pure water forms <u><em>hydroxide and hydronium ions.</em></u>

Explanation:

Ionization is a reaction in the pure water in which water breaks down into its constituting ions that hydronium ion and hydroxide ions.

H_2O+H_2O\rightleftharpoons H_3O^++OH^-

One molecule of water looses its proton to form hydroxide ion and l=the lost protons get associated with another water molecule to form hydronium ion.

4 0
3 years ago
Calculate the volumeof 1.0M NaOH necessary to completely neutralize 100mL of 0.50M of phosphoric acid (H3PO4)?
blondinia [14]

Answer:

50 mL

Explanation:

In case of titration , the following formula is used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ =  0.50M

V₁ = 100 mL

M₂ =  1.0M

V₂ = ?

Using the above formula , the volume of base , can be calculated as ,

M₁V₁ = M₂V₂

substituting the respective values ,

0.50M * 100 mL =  1.0M * V₂  

V₂  = 50 mL

3 0
3 years ago
An object with a mass of 15kg is accelerating at 10m/s2. What is the force carried by this object
sladkih [1.3K]

Answer: 150N

Explanation:

The formula to calculate the force will be the mass multiplied by the acceleration. In this case, mass = 15kg and acceleration = 10m/s². Therefore, force will be:

= Mass × Acceleration

= 15 × 10

= 150N

The force is 150Newton

8 0
3 years ago
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