Then, at a critical point during interphase (called the S phase), the cell duplicates its chromosomes and ensures its systems are ready for cell division. If all conditions are ideal, the cell is now ready to move into the first phase of mitosis.

8 0

3 years ago

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 at

ElenaW [278]

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$ .

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

4 0

3 years ago

2 Na + Cl2 → 2 NaCl

Ksivusya [100]

Approximately 71.77 grams. I think so

8 0

3 years ago

Read 2 more answers

The wavelength of the blue light given off by a mercury vapor street lamp is 457 nm. What is the frequency of this light in hert

Fed [463]

Answer:

6.56×10¹⁴ Hz

Explanation:

From the question given above, the following data were obtained:

Wavelength = 457 nm

Frequency =?

Next, we shall convert 457 nm to metre (m). This can be obtained as follow:

1 nm = 1×10¯⁹ m

Therefore,

457 nm = 457 nm × 1×10¯⁹ m / 1 nm

457 nm = 4.57×10¯⁷ m

Thus, 457 nm is equivalent to 4.57×10¯⁷ m

Finally, we shall determine the frequency of the blue light as follow:

Wavelength = 4.57×10¯⁷ m

Velocity of light = 3×10⁸ m/s

Frequency =?

Velocity = wavelength x frequency

3×10⁸ = 4.57×10¯⁷ × frequency

Divide both side by 4.57×10¯⁷

frequency = 3×10⁸ / 4.57×10¯⁷

frequency = 6.56×10¹⁴ Hz

Therefore, the frequency of the blue light is 6.56×10¹⁴ Hz

6 0

3 years ago