Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
<em>what should be ph-value to be strong acid?</em>
<em>what should be ph-value to be strong acid?</em><em>c</em><em>.</em><em> Less than </em><em>2</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
1.84 atm
Explanation:
P2=V1P1/V2
Temp is irrelevant because it is constant, so all you have to do is rearrange the ideal gas law.
You can check this by knowing P and V at constant T have an inverse relationship and therefore makes this the correct answer.
- Hope that helps! Please let me know if you need further explanation.
Answer:
The nervous system
Explanation:
The nervous system send messages from the brain to the muscles to move.
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
