How many significant figures are in 31400

Two students were conducting an experiment. They have two blocks of different metals and wanted to identify them. They started b

erik [133]

Answer:

this dosen't make sense

5 0

3 years ago

What action would you expect an atom of Aluminum to undergo?

KATRIN_1 [288]

Answer:

C. Lose three electrons to have a full outer shell

Explanation:

Al is in Group 13 of the Periodic Table, so it has three valence electrons.

It must either lose three electrons or gain five to achieve a stable octet.

It is easier to lose three electrons than it is to gain five, so Al loses three electrons.

D. is wrong, for the same reason.

A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.

C. is wrong. Al is a metal, so it will lose electrons in a reaction.

6 0

3 years ago

Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s

NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

$n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3$

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0

2 years ago

What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme

boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

4 0

3 years ago

In which step of a four-stroke engine cycle does the car release CO2, H2O, and CO?

In-s [12.5K]

Answer:

$\boxed{\text{exhaust stroke}}$

Explanation:

The four strokes are

Intake: The mixture of air and fuel enters the cylinder.
Compression: The mixture is compressed by a factor of about 11.
Ignition: A spark ignites the fuel, which burns and produces CO₂, CO, and H₂O.
Exhaust: The combustion products are expelled from the cylinder.

$\text{The cylinder releases CO$_{2}$, CO, and H$_{2}$O during the }\boxed{\textbf{exhaust stroke}}$

7 0

3 years ago