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Vedmedyk [2.9K]
4 years ago
7

Which is the first element in the periodic table to have an electron configuration in the 4th energy level?

Chemistry
2 answers:
Tju [1.3M]4 years ago
7 0
The answer is potassium:)
solong [7]4 years ago
4 0

The answer is potassium .......

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The temperature of their surroundings
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The main purpose of models is to gain a better understanding of the world true or false
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3 years ago
Why is saturated sodium chloride solution (also known as brine) used?
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6 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}

= 0.038 g

∴ Mass percentage of Nitrogen in the fertilizer = \frac{0.038 \ g}{0.568 \ g} * 100%

= 6.69%

8 0
4 years ago
Read 2 more answers
Calculate the displacement (the total volume of the cylinder through which the piston move) of a 5.70L automobile engine in cubi
Novay_Z [31]

Answer:

348 inches³

Explanation:

From our previous knowledge of  units conversion:

We know that 1000 cm³ makes 1 Liter.

Thus, for  a 5.70 L automobile engine in cubic meters will be:

= 5.70 × 1000 cm³

= 5700 cm³

Now, the displacement of the automobile in cubic inches provided that 1 inch = 2.534 cm is:

⇒ 5700× (1/ (2.54)³) in³

= 5700×0.0610 in³

= 347.7 in³

≅ 348 inches³

4 0
3 years ago
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