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2 years ago

A first-order reaction has a half-life of 20.0 minutes. Starting with 1.00 x 10^20 molecules

Chemistry

1 answer:

blsea [12.9K]2 years ago

7 0

Answer:

Explanation:

100 minutes is 5 half lives ( 100 min/20 min/half life)

= (1/2 )^5

1 x 10^20 * (1/2)^5 = 3.125 x10^18 molecules

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3 years ago

What is the bond dissociation energy for breaking all the bonds in a mole of o2 molecules?

djverab [1.8K]

<h3>Answer:</h3>

498 kj/mol

<h3>Explanation:</h3>

Chemical reactions occur as a result of bond breaking and bond formation.
The bonds in reactants are broken and atoms are rearranged to form new bonds.
During bond breaking energy is absorbed to break the bonds of reactants while bond formation involves the release of energy during the formation of new bonds.

In our case;

In 1 mole of the Oxygen molecule, there is one O=O bond

Energy absorbed to break O=O is 498 kJ/mol

Therefore, the ΔH required to break all the bonds in one mole of Oxygen(O₂) molecules is 498kJ/mol.

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3 years ago

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3 years ago

Read 2 more answers

An ideal gas is brought through an isothermal compression process. The 3.00 mol of gas goes from an initial volume of 261.6×10−6

Eduardwww [97]

Answer : The temperature and the final pressure of the gas is, 586.83 K and $1.046\times 10^{9}atm$ respectively.

Explanation : Given,

Initial volume of gas = $261.6\times 10^{-6}m^3$

Final volume of the gas = $138.2\times 10^{-6}m^3$

Heat released = -9340 J

First we have to calculate the temperature of the gas.

According to the question, this is the case of isothermal reversible compression of gas.

As per first law of thermodynamic,

$\Delta U=q+w$

where,

$\Delta U$ = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

$q=-w$

Thus, w = -q = 9340 J

The expression used for work done will be,

$w=nRT\ln (\frac{V_2}{V_1})$

where,

w = work done = 9340 J

n = number of moles of gas = 3 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = ?

$V_1$ = initial volume of gas

$V_2$ = final volume of gas

Now put all the given values in the above formula, we get the temperature of the gas.

$9340J=3mole\times 8.314J/moleK\times T\times \ln (\frac{261.6\times 10^{-6}m^3}{138.2\times 10^{-6}m^3})$

$T=586.83K$

Now we have to calculate the final pressure of the gas by using ideal gas equation.

$PV=nRT$

where,

P = final pressure of gas = ?

V = final volume of gas = $138.2\times 10^{-6}m^3=138.2\times 10^{-9}L$

T = temperature of gas = 586.83 K

n = number of moles of gas = 3 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get:

$P\times (138.2\times 10^{-9}L)=3mole\times (0.0821L.atm/mole.K)\times (586.83K)$

$P=1.046\times 10^{9}atm$

Therefore, the temperature and the final pressure of the gas is, 586.83 K and $1.046\times 10^{9}atm$ respectively.

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3 years ago

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of t

maks197457 [2]

Answer:

The gases will expand until a volume of Vf= 10812 cm^3

Explanation:

Since the gas mixture is expanding at constant pressure

$W=\int\limits^{Vf}_{Vi} {P} \, dV =P(Vf-Vi)$

and therefore

$Vf=W/P+Vi$

knowing that 1 torr=133,32 Pa

$p=635 torr * 133,322 Pa/torr * 1m^{3}/10^{6}cm^{3}=0,08466 J/ cm^{3}$

therefore

$Vf=912 J/(0,08466 J/cm^{3}) +40cm^{3} =10812 cm^{3}$

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3 years ago