Answer: A chlorine atom has an atomic number of 17 and a mass number of 38. This chlorine atom has 18 neutrons
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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:
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The relation between the volume of the gas and the temperature is established by Charles's law. With a decrease in the temperature, the volume decreases by 45.7 mL. Thus, option c is correct.
<h3>What is Charle's law?</h3>
Charle's law states the direct relation present between the temperature and the volume of the gas. The law is given as:
V₁ ÷ T₁ = V₂ ÷ T₂
Given,
V₁ = 50 mL
T₁ = 303.15 K
T₂ = 277.15 K
Substituting the value the final volume is calculated as:
50 ÷ 303.15 = V₂ ÷ 277.15
V₂ = (50 × 277.15) ÷ 303.15
= 45.71 mL
Therefore, option c. 45.7 mL is the final volume.
Learn more about Charles law here:
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Answer:
a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)
b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)
Explanation:
Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.
a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.
N₂O₅ + H₂O ⇒ HNO₃
Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.
N₂O₅ + H₂O ⇒ 2 HNO₃
b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.
Na₂O + H₂O ⇒ NaOH
Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.
Na₂O + H₂O ⇒ 2 NaOH
