Answer:
-586.56 kJ/mol is the standard enthalpy of the 3rd reaction.
Explanation:
...[1]
...[2]
..[3]
The unknown standard enthalpy of third reaction can be calculated by using Hess's law:
The law states that 'the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps'.
[2] - 2 × [1] = [3]
The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.The negative sign indicates that energy is released during this reaction.
Answer:
fraction of vacancies for this metal FV = 1.918*10⁻⁴
Explanation:
Given:
The number of vacancies per unit volume => ( Nv = 1*10²⁵ m⁻³ )
But we know that Avogrado's constant NA = 6.022*10²³ atoms/mol
Density of the material is given in g/cm3 we need to convert it to g/m³
Density of material ( p ) in g/m³ :
To convert we know that
1 g/cm³ = 1000000 g/m³ then
7.40 * ( 1000000 ) = 7.40*10⁶ g/m³
So, Density of material ( p ) in g/m³ = 7.40*10⁶ g/m³
Given Atomic mass = 85.5 g/mol
To Calculate the number of atomic sites per unit volume , we will use the below formula by substituting those values above
N = NA * p / A
N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5
N = 4.45*10³⁰ / 85.5
N = 5.212*10²⁸ atoms/m³
We can now Calculate the fraction of vacancies using the formula below;
Fv = Nv / N
Fv = 1*10²⁵ / 5.212*10²⁸
fraction of vacancies for this metal at 600c.= 1.918*10⁻⁴
Answer:
See below.
Explanation:
The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.
The balanced equation is:
2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)
From the equation, using atomic masses:
228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O
So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29
= 42.43 g of CO2.
Amount of water = 324.3 * 13.759) / 228.29
= 19.55 g of H2O.
Answer:
Tree sap flows over the leaf and preserves it.
Explanation:
Amber would preserve the image.
Answer:
Waves; wavelength; electromagnetic energy; ultraviolet light.
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases.
Lightwave can be defined as an electromagnetic wave that does not require a medium of propagation for it to travel through a vacuum of space where no particles exist.
Hence, sound and light are both found as waves, with a variety of wavelengths. The sun, a source of light waves specifically, releases a type of electromagnetic energy. It can be found as UVA or UVB types. These lights give off different levels of ultraviolet light, some of which can be harmful.
Additionally, the ultraviolet spectrum is divided into three categories and these are; UVA, UVB ,and UVC.
