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alexandr402 [8]
3 years ago
14

PLS HELP!!

Chemistry
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

from the 1st equation:

  4NH3                  4NO

  4 *(68)                 4*30

  1216                     X  mass of NO = 536.5 g

from the 2nd Equation

  2NO           2NO2

   2*30            2* 46

   536.5          x         mass of NO2 = 822.6 grams

from the 3rd Equation

  3NO2                         2HNO3

   3*(46)                           2* (63)

     822.6                             X                mass of nitric acid = 751.06 gram

b)  % yields = ( 96.2%* 91.3% *91.4%)= 80.3%

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ExtremeBDS [4]

<u>Answer 2 :</u> The given electronic configuration for a neutral atom of phosphorous in its ground state is incorrect.

Explanation :

A neutral atom of phosphorous has 15 electrons.

The given electronic configuration is incorrect.

The reason is, According to Aufbau principle, the electrons will be first filled in the sub-shell having lower orbital energy. As from the given configuration, 3p sub-shell has lower orbital energy than 4s sub-shell. So, the electrons will be filled in 3p sub-shell first. Hence, the ground state electronic configuration of neutral atom of phosphorous is,

1s^22s^22p^63s^23p^3

<u>Answer 3 :</u>

Element                     Rubidium              Magnesium                Aluminium

Symbol                             Rb                         Mg                              Al

Group number                  1                             2                               13

Number of valence          1                             2                                3

electrons

The order of general reactivity on the basis of number of valence electrons.

Rb > Mg > Al

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5 0
2 years ago
A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.
adell [148]

Answer:

a) V=4.3x10^3mL

b) V=4.3x10^6mm^3

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Explanation:

Hello,

a) In this case, the given height in cm is:

h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm

And the radius in cm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL

b) In this case, the given height in mm is:

h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm

And the radius in mm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm

Thus, the volume in cubic millimeters is:

V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\   \\\rho=1.5x10^5g/L

Best regards.

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