When an atom of N-14 is bombarded by an alpha particle, the single product is

Chemistry

1 answer:

puteri [66]3 years ago

6 0

Answer:

¹⁸F₉

Explanation:

Alpha particles contain 2 protons and 2 neutrons, that is 4/2.

Which mains that the number of protons and neutrons will increase by 2.

So, the single product will be ¹⁸F₉.

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An isotope with the combination of subatomic particles (69n,50p+,50e-) is?

faust18 [17]

Answer:

A typical atom consists of three subatomic particles: protons, neutrons, and electrons (as seen in the helium atom below). Other particles exist as well, such as alpha and beta particles (which are discussed below). The Bohr model shows the three basic subatomic particles in a simple manner. Most of an atom's mass is in the nucleus—a small, dense area at the center of every atom, composed of nucleons. Nucleons include protons and neutrons. All the positive charge of an atom is contained in the nucleus, and originates from the protons. Neutrons are neutrally-charged. Electrons, which are negatively-charged, are located outside of the nucleus.

Explanation:

4 0

3 years ago

Using 273.15 K as the value for Zero degree Celsius convert 45.1° C to Kelvin. 6.05 K 318 K 318.3 K 1.23 x 104 K

OLEGan [10]

Zero degree celcius = 273.15 degree kelvin

Simply, to get the value of 45.1 degree celcius in kelvin, we will add 273.15 to the given value (45.1).

Degree in kelvin = 45.1 + 273.15 = 318.25 degree kelvin

Approximating to the nearest tenth, the value will be 318.3 degree kelvin

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 $Al_2O_3$