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svp [43]
3 years ago
7

How many “p” orbital in principal energy level below?​

Chemistry
1 answer:
marshall27 [118]3 years ago
5 0

Answer:

Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. Therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

Explanation:

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Three samples of the same metal are weighed and their masses are found to be 44.40 g, 40.58 g, and 38.35 g. The corresponding vo
lidiya [134]

The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are, \frac{44.40}{4.8} = 9.25g/mL, \frac{40.58}{4.7} = 8.634g/mL and \frac{38.35}{4.2} = 9.130g/mL respectively.

The equation of the standard deviation is √{∑(x  - \frac{}{x})÷N}

Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.

The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.

The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.

The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.

The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be \sqrt{0.070}=0.265. Thus the standard deviation of the density value is 0.265g/mL.  

5 0
3 years ago
If 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen?
lyudmila [28]

The balanced reaction is:<span>

Mg (s) + 2HCl (aq) --> MgCl 2 (aq) + H 2 (g) 

We are given the amount hydrochloric acid to be used for the reaction. This will be the starting point of the calculation.

40.0 g HCl ( 1 mol HCl / 36.46 g HCl) (1 mol H2 / 2 mol HCl) (2.02 g H2 / 1 mol H2) = 1.11 g H2</span>

8 0
3 years ago
How many sig-figs in the following measurements?
DerKrebs [107]

Answer:

1.605cm = 4 significant figures

16.050cm = 4 significant figures

16.050cm = 4 significant figures

12 + 12.5 + 125 = 149.5 = 4 significant figures

1.62 × 10^3/2.8 × 10^-5 = 1620/0.000028 = 11 significant figures

4 0
3 years ago
how many grams of calcium oxide are producer from the decomposition of 125 grams of calcium carbonate
Dmitrij [34]

70.0 g. The decomposition of 125 g CaCO3 produces 700 g CaO.

MM = 100.09 56.08

CaCO3 → CaO + CO2

Mass 125 g

a) Moles of CaCO3 = 125 g CaCO3 x (1 mol CaCO3/100.09 g CaCO3)

= 1.249 mol CaCO3

b) Moles of CaO = 1.249 mol CaCO3 x (1 mol CaO/1 mol CaCO3)

= 1.249 mol CaO

c) Mass of CaO = 1.249 mol CaO x (56.08 g CaO/1 mol CaO) = 70.0 g

7 0
4 years ago
Using the atomic masses and relative abundance of the isotopes of nitrogen given below, determine the average atomic mass of nit
mylen [45]
The formula for average atomic mass is :

mass of isotope A * % of isotope A + mass of isotope B * % of isotope B + ....

Now,
Here,
Average atomic mass of nitrogen = (14.003 * 99.63%) + (15.000 * 0.37%)
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3 0
3 years ago
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