What is the element of 1s^22s^2p^63s^23p^64s^23d^104p^65s^24d^105p^1

A laboratory manual gave precise instructions for carrying out the reaction described by the chemical equation, C2H6O + O2 C2H4O

Alexxandr [17]

C2H6O + O2 ---> C2H4O2 + H2O

using the molar masses:-
24+ 6 + 16 g of C2H6O produces 24 + 4 + 32 g C2H4O2 (theoretical)
46 g produces 60g

60 g C2H4O2 is produced from 46g C2H6O
1g . .................................46/60 g
700g ................................. (46/60) * 700 Theoretically

But as the yield is only 7.5%

the required amount is ((46/60) * 700 ) / 0.075 = 7155.56 g

= 7.156 kg to nearest gram. Answer

8 0

2 years ago

A gas cylinder filled with nitrogen at standard temperature and pressure has a mass of 37.289 g. The same container filled with

andrew-mc [135]

Answer:

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

Explanation:

Let the moles of nitrogen gas = x moles

Moles of carbon dioxide = x moles ( As both are filled at same temperature and pressure conditions )

Given:

$Mass_{Container}+Mass_{Nitrogen\ gas}=37.289\ g$

Molar mass of nitrogen gas, $N_2$ = 28.014 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass}{28.014\ g/mol}$

Mass of nitrogen gas = 28.014x g

So,

Let, $Mass_{Container}=y$

$y+28.014x=37.289$

Similarly,

$Mass_{Container}+Mass_{Carbon\ dioxide\ gas}=37.440\ g$

Molar mass of nitrogen gas, $CO_2$ = 44.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$x\ moles= \frac{Mass{44.01\ g/mol}$

Mass of nitrogen gas = 44.01x g

So,

$y+44.01x=37.440$

Solving the two equations, we get :

$Mass_{Container}=y=37.025\ g$

x = 0.00943 moles

Thus, Given:

$Mass_{Container}+Mass_{Unknown\ gas}=37.062\ g$

$37.025\ g+Mass_{Unknown\ gas}=37.062\ g$

Mass of the gas = 0.037 moles

Moles = 0.00943 moles

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.00943\ moles= \frac{0.037\ g}Molar mass}$

Molar mass = 3.9236 g/mol ≅ 4 g/mol

This corresponds to Helium gas.

7 0

2 years ago

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

Aleksandr-060686 [28]

To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.

Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed

Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:

Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s

5 0

2 years ago

Read 2 more answers

Describe how a chemist should set up his or her stoichiometric calculation to predict the mass of CuSO4 that forms when a specif

DanielleElmas [232]

1) Calculate the number of moles of Cu SO4 . 5H20 by dividing the specified mass by the molar mass.

2) The ratio of production given by the equation is 1 mol of Cu SO4 . 5 H2O to 1 mol of Cu SO4=> 1:1, meaning that the number of moles of Cu SO4 produced is the same number of moles of Cu SO4.5H20 heated.

3) Finally mutiply the number of moles of Cu SO4 by its molar mass and there you have the mass of Cu SO4 produced.

5 0

2 years ago

Read 2 more answers

Describe the hybrid orbitals used by the central atom and the types of bonds formed in o3

Tems11 [23]

Hybridization in ozone, O3......

...O = O ........ 1 lone pair on central O, 2 lone pairs on terminal O 
../ 
O .................. 3 lone pairs on terminal O 

I didn't show the second of two resonance structures in which the single and double bonds are reversed. In reality, both bonds are identical have a bond order of 1.5 due to delocalized pi-bonding. 

The central atom exhibits sp2 hybridization since there is trigonal planar electron pair geometry. The notion of hybrid orbitals was "invented" by Linus Pauling in the 1930's as a way of explaining the geometry of molecules, primarily the geometry of carbon compounds. 

If the electron pair geometry is linear, the hybridization is sp. 
If the electron pair geometry is trigonal planar, the hybridization is sp2. 
If the electron pair geometry is tetrahedral, the hybridization is sp3. 

The notion that there is sp3d and sp3d2 because of d-orbital participation has been debunked. Chemists know today that there is no d-orbital involvement in hypervalent molecules regardless of what some out-of-date textbooks and some teachers' dusty old notes may say. Instead, the best explanation involves 3-center, 4-electron bonding.

8 0

2 years ago