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3 years ago

Strong bases such as NaOH follow the ionic IUPAC naming convention. Predict the name of this ternary compound.

Chemistry

2 answers:

GalinKa [24]3 years ago

8 0

Answer:

Sodium Hydroxide

Explanation:

Nitella [24]3 years ago

7 0

Answer:Hence, the name of the NaOH is Sodium hydroxide.

Explanation:

In strong base NaOH is made up of three element sodium , oxygen and hydrogen.

NaOH dissociates is into cation and anion in its aqueous solution:

$NaOH(aq)\rightarrow Na^++OH^-$

While naming the ternary compound:

First write the name of cation present, here the cation is of sodium that $Na^+$ .
followed by space then write name of the anion , here anion present in a polyatomic anion that $OH^-$ , the name of this polyatomic anion is hydroxide.

Hence, the name of the NaOH is Sodium hydroxide.

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marishachu [46]

Answer:

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Explanation:

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Oxidation:

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Reduction:

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3 years ago

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4 years ago

Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Hf = –393.5 kJ/mol), and H2O(g) (Hf = –241.8

Artyom0805 [142]

Answer:

$\boxed{\text{(A) -2657.4 kJ/mol}}$

Explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

2C₄H₁₀ g) + 13O₂(g) ⇌ 8CO₂(g) + 10H₂O(g)

ΔH°f/kJ·mol⁻¹: -125.7 0 -393.5 -241.82

$\begin{array}{rcl}\Delta_{\text{c}}H^{\circ} & = & [8(-393.5) + 10(-241.82)] - 2(-125.7)\\& = & [-3148.0 - 2418.2] +251.4\\& = & -5566.2 + 251.4\\& = & -5314.8\\\end{array}\\\\\text{This is the value for 2 mol of butane.}\\\text{The enthalpy of combustion is } \dfrac{-5314.8}{2} = \boxed{\textbf{-2657.4 kJ/mol}}$

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3 years ago