Answer:
58mL
Explanation:
Given parameters:
Density of water = 1g/mL
Mass of object = 58g
Unknown:
The volume the object must have to be able to float in water = ?
Solution:
To solve this problem, we know that the object must have density value equal to that of water or less than that of water to be able to float.
We then set its density to that of water;
Density = 
Volume = 
So;
Volume = 
= 58mL
Answer:
100 %
Explanation:
The maximun efficiency possible (whem not limited by the second law of thermodynamics) happens when all the energy used is transformed into the type of energy we required with no other transformations.
For example, in an engine we want that all the energy we supply is being converted to work. That's the ideal case, but in reality always some of that energy is lost in the form of heat.
Answer:
Wet ashing is primarily used in the preparation of samples for subsequent analysis of specific minerals . It breaks down and removes the organic matrix surrounding the minerals so that they are left in an aqueous solution
Explanation:
Wavelength is 6.976 x 10^ -35 m
Explanation:
In this, we can use De Broglie’s equation. This equation is the relationship between De Broglie’s wavelength, velocity and the mass of a moving object. In this equation, we are using plank's constant which is 6.626 x 10^-34 m^2 kg/s.
We know that one mile per hour is equivalent to 0.447 M/S.
And One gram is equivalent to 10^-3 kg.
De Broglie’s wavelength = λ ( wave length) = Plank’s constant/ Mass x velocity
λ ( wave length) = 6.626 x 10^ -34/ (425 x10^-3) x ( 50 x 0.447)
= 6.626 x 10^ -34/ 0. 425 x 22.35
= 6.626 x 10^ -34/ 9.498
= 6.976 x10^ -35 m
So, the wavelength of the football will be 6.976 x 10^ -35 m
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
