when the metal is lost heat and the calorimeter of water is gained the heat
and when the heat lost = the heat gained so,
(M*C*ΔT)m = (M*C*ΔT)w
when Mm= mass of the metal = 30 g
Δ Tm = (80-25) = 55 °C
and Mw = mass of water = 100 g
Cw is the specific heat of water = 4.181 J/g.°C
ΔTw = (25-20) = 5 °C
so by substitution:
∴ 30* Cm*55 = 100 * 4.181 * 5
∴Cm (specific heat of metal) = (100*4.181*5)/(30*55)
∴C of metal = 1.267 J/g.°C
Answer:
The enthlapy of solution is -55.23 kJ/mol.
Explanation:
Mass of water = m
Density of water = 1 g/mL
Volume of water = 50.0 mL
m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g
Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C
Heat capacity of water,c =4.186 J/g°C
Heat gained by the water when an unknown compound is dissolved be Q
Q= mcΔT
heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.
Q'=-Q
Q'= -983.71 J =-0.98371 kJ
Moles of unknown compound = 
The enthlapy of solution :
The enthlapy of solution is -55.23 kJ/mol.
Answer:
See explanation
Explanation:
Hydrogen has a valency of +1 or -1. Its electronic configuration is 1s1.
The 1s sub-level (first shell) is known to hold two electrons. This means that hydrogen may either loose this one electron in the 1s level to yield H^+ or accept another electron into this 1s level to form H^- (the hydride ion).
The formation of the hydride ion completes the 1s orbital.
Answer:
59.2 grams
Explanation:
We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:
There are 59.2 grams of water in this solution.
Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:
<span>H2C=O---------H-OH </span>
<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>
<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>
<span>H2C=O + H2O <-----------------> H2C(OH)2</span>
