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Oduvanchick [21]
3 years ago
7

Select all that apply.

Chemistry
1 answer:
inysia [295]3 years ago
8 0

Answer: has zero mass

is an electron

has a -1 charge

is electromagnetic energy

Explanation:

A beta-particle is an electron emitted from the nucleus of an atom following the decay of a neutron into a proton. Beta rays are deflected by a magnetic field in a direction that indicates a negative charge, thus they are electromagnetic in nature.

The beta particle released is basically a electron with -1 charge and no mass.

General representation of beta particle in the form of _Z^A\textrm {X} given as _{-1}^0e

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

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When 0.100 mol of carbon is burned in a closed vessel with8.00
antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of O_2 = 8.00 g

Molar mass of O_2 = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of O_2.

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C+O_2\rightarrow 2CO

From the balanced reaction we conclude that

As, 2 mole of C react with 1 mole of O_2

So, 0.1 moles of C react with \frac{0.1}{2}=0.05 moles of O_2

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

Now we have to calculate the moles of CO

From the reaction, we conclude that

As, 2 mole of C react to give 2 mole of CO

So, 0.1 moles of C react to give 0.1 moles of CO

Now we have to calculate the mass of CO

\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO

\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g

Therefore, the mass of carbon monoxide form can be 2.8 grams.

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Answer:

Explanation:

I think option d is correct

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<u />

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