An alkoxide is an organic functional group formed when a hydrogen atom is removed from a hydroxyl group of alcohol when reacted with a metal. It is the conjugate base of alcoho.
Answer:
Q= mc∆T(ice) + mLF(ice) + mc∆T(water) + mLV(water) + mc∆T(steam)
m=250 g = 0.25 kg = ¼ kg c(ice)= 2100 J/kg.K c(water)= 4200 J/kg.K LF(ice)= 333.7 kJ/kg LV(water)= 2256 kJ/kg c(steam)= 2080 J/kg.K
Explanation:
Q= ¼ × 2100 × (0°-(-30°)) + ¼ × 333700 + ¼ × 4200 × (100°-0°) + ¼ × 2256000 + ¼ × 2080 × (145°-100°)
Q= 15750 + 83425 + 105000 + 564000 + 23400
Q= 791575 J
Here you are! I hope it helps, and also for the ones I put a red ‘x’ it depends on how you round it.
Answer:
A. 0.000128 M is the solubility of M(OH)2 in pure water.
B. 
is the solubility of 
in a 0.202 M solution of 
.
Explanation:
A
Solubility product of generic metal hydroxide = 
S 2S
The expression of a solubility product is given by :
![K_{sp}=[M^{2+}][OH^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BM%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Solving for S:
0.000128 M is the solubility of M(OH)2 in pure water
B
Concentration of = 0.202 M
Solubility product of generic metal hydroxide =
S 2S
So, ![[M^{2+}]=0.202 M+S](https://tex.z-dn.net/?f=%5BM%5E%7B2%2B%7D%5D%3D0.202%20M%2BS)
The expression of a solubility product is given by :
Solving for S:
is the solubility of in a 0.202 M solution of .
The standard enthalpy of reaction of the given reaction is -865.71 kJ per mole of N₂H₃CH₃.
<h3>What is the standard molar enthalpy of formation?</h3>
The standard molar enthalpy of formation of a compound is defined as the enthalpy of formation of 1.0 mol of the pure compound in its stable state from the pure elements in their stable states at P = 1.0 bar at a constant temperature.
Let's consider the following equation.
4 N₂H₃CH₃(l) + 5 N₂O₄(l) → 12 H₂O(g) + 9 N₂(g) + 4 CO(g)
We can calculate the standard enthalpy of the reaction using the following expression.
ΔH° = Σnp × ΔH°f(p) - Σnr × ΔH°f(r)
where,
- ΔH° is the standard enthalpy of the reaction.
- n is stoichiometric coefficient.
- ΔH°f is the standard molar enthalpy of formation.
- p are the products.
- r are the reactants.
ΔH° = 12 mol × ΔH°f(H₂O(g)) + 9 mol × ΔH°f(N₂(g)) + 4 mol × ΔH°f(CO(g)) - 4 mol × ΔH°f(N₂H₃CH₃(l)) - 5 mol × ΔH°f(N₂O₄(l))
ΔH° = 12 mol × (-241.81 kJ/mol) + 9 mol × (0 kJ/mol) + 4 mol × (-110.53 kJ/mol) - 4 mol × (54.20 kJ/mol) - 5 mol × (-19.56 kJ/mol)
ΔH° = -3462.84 kJ
In the balanced equation, there are 4 moles of N₂H₃CH₃. The standard enthalpy of reaction per mole of N₂H₃CH₃ is:
-3462.84 kJ / 4 mol = -865.71 kJ/mol
The standard enthalpy of reaction of the given reaction is -865.71 kJ per mole of N₂H₃CH₃.
Learn more about enthalpy here: brainly.com/question/11628413
