1. I think it is true?
2. Low melting points
3. True
4. Atomic number, I think it’s periods?
5. Groups?
Sorry, I might not get all of them right :(
Hope this helps you in any way!!
If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
If you search the question text (starting with "Write") you will definitely find the answer
Answer:
everything including the space...
hope it helps...
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
