Please select the word from the list that best fits the definition

Chemistry

2 answers:

Vera_Pavlovna [14]3 years ago

6 0

I believe it is c

Explanation:

because if u think about it and also do some research you would see that the circulatory system is a strong part of your body which can help u through natural defences if this sounds weird it's all in research but if it ain't c dont blame me for ruining your life- lol but yeah I think its c

AveGali [126]3 years ago

3 0

Answer:

I choose D option because may be it's correct

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A 0.568-g sample of fertilizer contained nitrogen as ammonium sulfate, . It was analyzed for nitrogen by heating with sodium hyd

Mariulka [41]

Answer:

6.69%

Explanation:

Given that:

Mass of the fertilizer = 0.568 g

The mass of HCl used in titration (45.2 mL of 0.192 M)

= $0.192*\frac{45.2}{1000}* \frac{36.5}{1 \ mole}$

= 0.313 g HCl

The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)

= $\frac{44.3 \ mL * 1.0 \ L}{1000 \ mL} *0.133 \ mole/L$

= 0.0058919 mole of NaOH

From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process

Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g

= 0.2151 g HCl

From above ; the total amount of HCl used = 0.313 g

The total amount that is used for complete neutralization = 0.2151 g

∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g

= 0.0979 g

We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g

Now; the amount of HCl neutralized by 0.0979 HCl = $\frac{17}{36.5}*0.0979$

= 0.0456 g

Therefore, the mass of nitrogen present in the fertilizer is:

= $0.0456 \ g \ NH_3 * \frac{1 \ mol \ NH_3 }{17.0 \ mol \ of \ NH_3} * \frac{1 \mol \ (NH_4)_2SO_4}{2 \ mol \ NH_3 } * \frac{2 \ mol \ N }{1 \mol \ (NH_4)_2SO_4}* \frac{14.0 g }{1 \ mol \ N}$