What element has 4 energy levels and 8 valence electrons?

Other two substances present in breathed out air <br>

Angelina_Jolie [31]

Answer:

Gases we Breathe Out

It is the same air that we inhale. ... The amount of inhaled air contains 21% of oxygen and 0.04% of carbon dioxide, while the air we breathe out contains 16.4% of oxygen and 4.4% of carbon dioxide.

6 0

3 years ago

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Substances like neon , which is a gas at room temperature and pressure, can often be liquified or solidified only at very low te

marshall27 [118]

Answer:

Condenses at 27.25K.

Freezes at 24.65K.

Explanation:

In order to solve this above question, there is is need to make use of the following equation. The main idea here is to convert degree celsius to Kelvin. Hence,

0°C + 273.15 = 273.15K---------------------(1).

Therefore, we will make use of the above equation (1) and slot in the values for at degree celsius at which it condenses and at degree celsius at which it freezes.

So, we have at temperature at which it condenses:

-245.9°C + 273.15 = 27.25K.

Also, we have at temperature at which it freezes.

-248.5°C + 273.15 = 24.65K.

8 0

3 years ago

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:

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