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:
Answer:
Metallic character decreases, and electronegativity increases.
Explanation:
Hello!
In this case, according to the organization of the periodic table, we can see that from left to right, the electronegativity increases as nonmetals are able to attract electrons more easily than metals.
Moreover, in contrast to the previous periodic trend, the metallic character decreases from left to right because the elements tend to decrease the capacity to lose electrons and consequently start attracting them.
Thus, the answer would be: "Metallic character decreases, and electronegativity increases".
Best regards!
57.0 is it rounded to three sig figs. You count three spaces then round from there, which would be the zero and you would round down because the four is there.
Answer:
P = 27.9 atm
Explanation:
Given data:
Mass of CO₂ = 25 g
Temperature = 25°C (25+273.15 K = 298.15 K)
Volume of gas = 0.50 L
Pressure of gas = ?
Solution:
Firs of all we will calculate the number of moles of gas,
Number of moles = mass/molar mass
Number of moles = 25 g/ 44 g/mol
Number of moles = 0.57 mol
Pressure of gas :
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
P × 0.50 L = 0.57 mol × 0.0821 atm.L/ mol.K × 298.15 K
P = 13.95 atm.L/ 0.50 L
P = 27.9 atm
