An early model of the atom was developed in 1913 by Danish scientist Niels Bohr (1885–1962). The Bohr model shows the atom as a central nucleus containing protons and neutrons with the electrons in circular orbitals at specific distances from the nucleus . These orbits form electron shells or energy levels, which are a way of visualizing the number of electrons in the various shells. These energy levels are designated by a number and the symbol "n." For example, 1n represents the first energy level located closest to the nucleus.

6 0

3 years ago

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago

combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

6 0

3 years ago