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myrzilka [38]
3 years ago
9

What gas law states that pressure and temperature are directly proportional at

Chemistry
2 answers:
Rus_ich [418]3 years ago
8 0
Charles Law is the law that states pressure and temperature are directly proportional at constant volume and miles
777dan777 [17]3 years ago
3 0
Charles law is the gas law
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How many grams do 1.5 x 1026 atoms of magnesium weigh?
LUCKY_DIMON [66]

Answer:

0.3

Explanation:

if I'm wrong don't mark it against me

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3 years ago
A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon
svetlana [45]

Answer:

\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}

Explanation:

1. Miles travelled in an average month  

\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}

2. Using a gasoline powered vehicle  

(a) Moles of heptane used  

n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}

(b) Equation for combustion  

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

(c) Moles of CO₂ formed  

n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}

(d) Volume of CO₂ formed  

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.  

V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}

3. Using an electric vehicle  

(a) Theoretical energy used  

\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}

(b) Actual energy used  

The power station is only 85 % efficient.  

\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

The heat of combustion of methane is -802.3 kJ·mol⁻¹  

V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}

4. Comparison  

\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}

6 0
3 years ago
*WILL GIVE BRAINLIEST TO CORRECT ANSWER *<br> *QUESTION IS ON PICTURE*
Ivahew [28]

Answer:

A

Explanation:

When a temperature increases particles tend to collide with each other often as they gain kinetic energy making them collide much more often and they'll collide with more energy due to the increase of kinetic energy

7 0
3 years ago
Read 2 more answers
An aqueous solution containing 10 g of an optically pure compound was diluted to 500 mL with water and was found to have a speci
lapo4ka [179]

Answer:

<u>Optical purity = 76.9231 %</u>

<u>Specific rotation of mixture = - 97.6923 °</u>

Explanation:

The mass of the racemic mixture = 3 g

It means it contains R enantiomer = 1.5 g

S enantiomer = 1.5 g

Amount of Pure R = 10 g

Total R = 11.5 g

Total volume = 500 mL + 500 mL = 1000 mL = 1 L

[R] = 11.5 g/L

[S] = 1.5 g/L

Enantiomeric excess = \frac {Excess}{Total\ Concentration}\times 100 = \frac {11.5-1.5}{11.5+1.5}\times 100 = 76.9231 %

<u>Optical purity = 76.9231 %</u>

Also,

Optical purity = \frac {optical\ rotation\ of\ mixture}{optical\ rotation\ of\ pure\ enantiomer}\times 100

Optical rotation of pure enantiomer = −127 °

76.9231=\frac {optical\ rotation\ of\ mixture}{-127^0}\times 100

<u>Specific rotation of mixture = - 97.6923 °</u>

6 0
3 years ago
What causes the changes in air pressure on the earth surface
Fudgin [204]
<span>Atmospheric Pressure</span>
5 0
3 years ago
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