Answer:
Since groups and clusters contain so many galaxies relatively close together, it should not be surprising that galaxies sometimes collide with each other. Although galaxy collisions are common, stars in each galaxy are so far apart that collisions between stars are very rare
Co2 is made of carbon ( atomic mass: 12) and oxygen (atomic mass : 16)
First : mass of 1 C atom + mass of 2 O atom = 12+(16*2)=44 amu so 1 mole CO2 =44 grams then 2.5 mole CO2 =44*2.5 =110 grams
So the answer is 110 grams
Answer:
A.) because they have delocalized electrons
Explanation:
This is right on edg. 2020
Hello there!
From my calculations the answers are:
D. Mass
B. Gravity
B. Gravity
Hope This Helps You!
Good Luck :)
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: