Answer:Tungsten changes oxidation numbers +6 to zero
Undergoes reduction
Explanation:
did it!
True, a lunar eclipse occurs when the Moon is between the Sun and Earth
Answer:
Noble Gases
Explanation:It's not on there but Noble gases is the least reactive element because it has 8 electrons and their outer energy is full.
Answer: The heat required is 6.88 kJ.
Explanation:
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of ethanol = 25.0 g
= specific heat of solid ethanol= 0.97 J/gK
= specific heat of liquid ethanol = 2.31 J/gK
n = number of moles of ethanol = 
= enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole
= change in temperature
The value of change in temperature always same in Kelvin and degree Celsius.
Now put all the given values in the above expression, we get
![\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B25.0%20g%5Ctimes%200.97J%2FgK%5Ctimes%20%28-114-%28-135%29K%5D%2B0.534mole%5Ctimes%205020J%2Fmole%2B%5B25.0g%5Ctimes%202.31J%2FgK%5Ctimes%20%28-50-%28-114%29%29K%5D)
(1 KJ = 1000 J)
Therefore, the heat required is 6.88 kJ
Answer:
Fe =52.2%, O = 44.9%, H = 2.81%
Explanation:
Percentage composition is also known as percentage by mass.
First, find the Mr of the compound.
Mr of Fe(OH)₃ = 55.8 + (16 × 3) + (1 × 3)
= 106.8
Now Divide the mass of the element, as per the compound, by the Mr of the compound found and times that by 100.
% composition of Fe =
× 100 = 52.2%
% composition of O =
× 100 = 44.9%
[There are 3 oxygen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of oxygen with the number of atoms in the compound: 16×3 ]
% composition of H =
× 100 = 2.81%
[There are 3 hydrogen atoms in the compound Fe(OH)₃, so we will multiply the atomic mass of hydrogen with the number of atoms in the compound: 1×3 ]