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MrMuchimi
3 years ago
11

Calculate the density (in g/l) of ch4(g) at 75 c and 2.1 atm. (r = 0.08206 latm/molk

Chemistry
1 answer:
vaieri [72.5K]3 years ago
4 0
Answer is: density of methane is 1.176 g/L.<span>
V(CH</span>₄<span>) = 1 L.
T = 75°C = 348.15 K.
p = 2.1 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
n<span> = p·V / R·T.
n</span></span>(CH₄) = 2.1 atm · 1 L / 0.08206 L·atm/mol·K · 348.15 K.
n(CH₄) = 0.0735 mol.
m(CH₄) = n(CH₄) · M(CH₄).
m(CH₄) = 0.0735 mol · 16 g/mol.
m(CH₄) = 1.176 g.
d(CH₄) = m(CH₄) ÷ V(CH₄).
d(CH₄) = 1.176 g/L.
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D. 3 miles (4.5 kilometers)

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3 years ago
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The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into H_{2}, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.

An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as \pi* orbitals.

Therefore,  molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>

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Hence, the correct answer will be option (b)

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To know more about molecular orbital

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6 0
2 years ago
Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h
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<u>Answer:</u>

<em>0.264 g of CO_2 can be formed from 288 mg of O_2</em>

<u>Explanation:</u>

The balanced chemical equation is

2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O

The conversions are  

Mass in mg O_2 is converted to mass in g O_2  

Mass in g O_2 is converted to moles O_2 by dividing with molar mass  

Moles O_2 is converted to moles CO_2  by using the mole ratio of O_2:CO_2 is 9 : 6

Moles CO_2  is converted to mass CO_2 by multiplying with molar mass CO_2

mass in mg O_2  > mass in g O_2 >moles O_2 > moles CO_2 > mass CO_2

288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}

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-As the amount of carbon dioxide in the atmosphere increases, the oceans absorb a lot of it. In the ocean, carbon dioxide reacts with seawater to form carbonic acid which causes the acidity of seawater to increase.

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