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Kazeer [188]
3 years ago
14

If you could measure the mass of the substances inside the cup before and after the reaction, you would see that the mass decrea

sed over a course of the reaction. Explain the difference.
Chemistry
2 answers:
Shalnov [3]3 years ago
8 0
Gases may have formed during the reaction, released into the air, thus lowering the mass after the reaction of the remaining products.
swat323 years ago
4 0
The difference is the mass of the gases that were formed during the reaction. If the cup is open, it is always the case that the gas would escape. Any reaction that forms a gas as a product is called a gas forming reaction. 
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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
The reactants in the above reaction are hydrogen and nitrogen. The product is ammonia. What is a balanced equation for that
Elanso [62]
N2 + 3H2 (arrow) 2NH3
4 0
3 years ago
An equimolar mixture of acetone and ethanol is fed to an evacuated vessel and allowedto come to equilibrium at 65°C and 1.00 atm
frutty [35]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

3 0
3 years ago
Mercury thermometers work because mercury ____ when it is warmed
kap26 [50]
Mercury expands when it is heated. This process is called thermal expansion.
5 0
3 years ago
If a 250mL beaker weighs 13.473g and the same beaker plus 2.2 mL of water weighs 15.346g. How much does the 2.2 mL of water, alo
Sindrei [870]

Given :

A 250 ml beaker weighs 13.473 g .

The same beaker plus 2.2 ml of water weighs 15.346 g.

To Find :

How much does the 2.2 ml of water, alone, weigh .

Solution :

Now, mass of water is given by :

\text{Mass of water alone = Total mass - Mass of beaker alone}\\\\Mass =15.346-13.473\ g\\\\Mass =1.873\ g

Therefore , mass of 2.2 ml of water alone is 1.873 g .

Hence , this is the required solution .

4 0
3 years ago
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