Answer:
Explanation:Puesto que la energía causante de la aceleración del espacio-tiempo no ha podido ser observada en forma directa, se ha dado en llamarla energía oscura. Dos candidatos teóricos que podrían hacer las veces de esta energía son una constante cosmológica no igual a cero (que pudo haber causado la inflación cósmica) y una energía repulsiva más general llamada quintaesencia.
Answer:
oh i love football. i know this. The guy with 200 kilograms because he is heavier o that means more weight going against the guy. Is this American football. cause i love American football
Explanation:
Potassium carbonate reacts with barium chloride to precipitate insoluble barium salts that is barium carbonate, when the solutions are mixed together you’ll see white precipitate that is barium carbonate and potassium is dissolve in the solution to form potassium chloride. Potassium chloride is soluble in water so you will not see solid potassium chloride precipitate out.
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Volume
Temperature
Amount of gas (miles)
Type of gas
Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:
