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max2010maxim [7]
3 years ago
7

On another planet, the isotopes of titanium have the following natural abundances.

Chemistry
1 answer:
anygoal [31]3 years ago
8 0

Answer:

Average atomic mass = 46.91466 amu

Explanation:

Step 1: Data given

Isotopes of titanium

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10.000 % ⇒ 47.94795 amu

50Ti = 19.100 % ⇒ 49.94479 amu

Step 2: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

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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
What has 20 electrons in 44 Mass
nadya68 [22]

It would be Calcium: Ca

6 0
3 years ago
Read 2 more answers
The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t
Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
What is difference in reaction of O2 with metal and nonmetal?​
aksik [14]

Answer:

Metal more reactive than non metal

4 0
3 years ago
What is the mass of KOH found in a 785 mL of a 1.43 M solution of KOH
My name is Ann [436]

Molarity is defined as the ratio of number of moles to the volume of solution in litres.

The mathematical expression is given as:

Molarity = \frac{Number of moles}{volume of solution in litres}

Here, molarity is equal to 1.43 M and volume is equal to 785 mL.

Convert mL into L

As, 1 mL = 0.001 L

Thus, volume = 785\times 0.001 = 0.785 L

Rearrange the formula of molarity in terms of number of moles:

Number of moles =molarity\times volume of solution in litres

n = 1.43 M \times 0.785 L

= 1.12255 mole

Now,  Number of moles  = \frac{mass in g}{molar mass}

Molar mass of potassium hydroxide  = 56.10 g/mol

1.12255 mole  = \frac{mass in g}{56.10 g/mol}

mass in g = 1.12255 mole\times 56.10 g/mol

= 62.97 g

Hence, mass of KOH = 62.97 g



7 0
3 years ago
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