Which of the following is not a base? A: NH3 B. NaOH C. H3C6H5O7 D. KOH

Chemistry

2 answers:

olga2289 [7]2 years ago

7 0

I think that C is the answer.....

irina1246 [14]2 years ago

7 0

Answer: Option (C) is the correct answer.

Explanation:

A base is a substance or specie that contains a lone pair of electron or a hydroxyl group, that is, OH group.

For example, $NH_{3}$ , NaOH, and KOH are all bases.

But $H_{3}C_{6}H_{5}O_{7}$ is not a base because there is no lone pair of electron and no hydroxyl group.

Thus, we can conclude that out of the given options $H_{3}C_{6}H_{5}O_{7}$ is not a base.

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

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In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$