The answer is H₃C₆H₅O₇ (OPTION D).

In essence, the question is asking which of the options is an acid or a neutral compound; thus not a base. A base is a compound that accepts free hydrogen ions while in aqueous solution and as such they usually lack hydrogen ions that can be released in solution. If you ionize all the compounds/molecules above, the only on that has free hydrogen ions is OPTION D (which is an organic acid and not a base).

3 0

3 years ago

The oxidation of ammonia produces nitrogen and water via the following reaction: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l) Suppose the

Sonbull [250]

Answer:

The rate of consumption of $NH_{3}$ is 2.0 mol/L.s

Explanation:

Applying law of mass action to this reaction-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

where $-\frac{\Delta [NH_{3}]}{\Delta t}$ represents rate of consumption of $NH_{3}$ , $-\frac{\Delta [O_{2}]}{\Delta t}$ represents rate of consumption of $O_{2}$ , $\frac{\Delta [N_{2}]}{\Delta t}$ represents rate of formation of $N_{2}$ and $\frac{\Delta [H_{2}O]}{\Delta t}$ represents rate of formation of $H_{2}O$ .

Here rate of formation of $H_{2}O$ is 3.0 mol/(L.s)

From the above equation we can write-

$-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

Here $\frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))$

So, $-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}$

Hence, $-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)$

6 0

3 years ago

In a Lewis diagram for methane (CH4), which atom or atoms is inside, or central?

Ahat [919]

Answer:

B carbon

Explanation

Lewis structure or dot structure is an easy way to get the bonding details of atoms in a molecule. If we talk about methane molecule carbon is central atom with four electrons that are bonded to four hydrogen atoms and each bond is single covalent bond.

Please see attached figure,

Hope it helps!

7 0

3 years ago

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