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1 year ago

7.The following symbol is used to represent the particles in what type of radiation?Select one:a. Alphab. Betac. Gammad. Neutron

Chemistry

1 answer:

laiz [17]1 year ago

5 0

Explanation:

The emission of a beta particle is the result of the rearrangement of the unstable nucleus of the radioactive atom in order to acquire stability. For that, a phenomenon occurs in the nucleus, in which a neutron decomposes giving rise to three new particles: a proton, an electron (β particle), and a neutrino. The antineutrino and electron are emitted. The proton, however, remains in the nucleus.

The symbol is used to represent beta particles.

Answer: b. Beta

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The answer for the following problem is described below.

Therefore the standard enthalpy of combustion is -2800 kJ

Explanation:

Given:

enthalpy of combustion of glucose(Δ $H_{f}$ of $C_{6}H_{12} O_{6}$ ) =-1275.0

enthalpy of combustion of oxygen(Δ $H_{f}$ of $O_{2}$ ) = zero

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standard enthalpy of combustion

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Δ $H_{f}$ = ∈Δ $H_{f}$ (products) - ∈Δ $H_{f}$ (reactants)

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Answer:

They would all exhibit the same pressure.

Explanation:

Since the same number of mole of each gas is placed in different containers, it means the gas will occupy the same volume.

Now, the gases were observed at the same temperature. This means they will all have the same pressure as their volume is the same.

Now we can further understand this by doing a simple calculation as follow:

Assumptions:

For H2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, H2 has a pressure of 1 atm.

For N2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, N2 has a pressure of 1 atm

For O2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, O2 has a pressure of 1 atm

For He:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, He has a pressure of 1 atm.

From the above illustrations we can see that the gases have the same pressure since they have the same number of mole, volume and were observed at the same temperature.

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