Answer:
The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.
Explanation:
Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;
P ∝ V
P V = K
P1 V1 = P2 V2
Parameters :
P1 = Initial pressure of the gas = 750 mmHg
V1 = Initial pressure of the gas = 2. 00 Liters
P2 = Final pressure of the gas = 600 mmHg
V2 = Fimal volume of the gas = ? Liters
Calculations :
V2 = P1 V1 ÷ P2
V2= 750 × 2. 00 ÷ 600
V2 = 1500 ÷ 600
V2 = 2.5 Liters.
Therefore, the new volume of the gas is 2. 5 Liters.
<u>Given: </u>
Radius of culvert, r = 0.5 m
Tangential acceleration of the truck, a = 3 m/s2
<u>To determine:</u>
The angular acceleration, α
<u>Explanation:</u>
The tangential acceleration is related to the angular acceleration through the radius as:
a = rα
α = a/r = 3 ms⁻²/0.5 m = 6 s⁻²
Ans: The angular acceleration is 6 s⁻²
Answer:
B.
Explanation:
electrons can be lost by one particle, and gained by another particle
The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
![\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol](https://tex.z-dn.net/?f=%20%5CDelta%20H%20%3D%20%5B2%2A%286%2A413%20%2B%20347%29%20%2B%207%2A498%20-%20%284%2A2%2A799%20%2B%206%2A2%2A467%29%5D%20kJ%2Fmol%20)
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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