5.30x10^22 x 261.35g/6.02x10^23 = 23.01 g Ba(NO3)2
Answer:
D) burning a candle
A demonstration that contains both physical and chemical changes will be BURNING OF A CANDLE.
<em>This is because when a candle is lighted up, on the top of it (where we see flame) the oxygen and other gases present around will be chemically changed (burned up)</em>
<em>When the wax in the top of the candle will get melted due to the heat of the fire a physical change takes places.</em>
Answer:
K = [HI]² / [H₂] [I₂]
Explanation:
To write the expression of equilibrium constant, K, it is important that we know how to obtain the equilibrium constant.
The equilibrium constant, K for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient. Thus, the equilibrium constant is written as follow:
K = [Product] / [Reactant]
Now, we shall determine the equilibrium constant for the reaction given in the question above. This can be obtained as illustrated below:
H₂(g) + I₂(g) —> 2HI (g)
K = [HI]² / [H₂] [I₂]
1 - 8
2 - 13
3 - 20
4 - 19
5 - 19
6 - 22
* protons - Atomic number
electrons - number of electrons is equal to number of protons k-41 is neutral
neutrons - Mass number - Atomic number
= 41 -19
= 22
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
<em>Where q is heat of reaction in J,</em>
<em>m is the mass of the solution in g,</em>
<em>S is specific heat of the solution = 4.184J/g°C</em>
<em>ΔT is change in temperature = 11.21°C</em>
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
