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Kryger [21]
3 years ago
15

Why are valence electrons important?

Chemistry
2 answers:
Usimov [2.4K]3 years ago
5 0
They are the outer layer of the electron layers.
frosja888 [35]3 years ago
5 0
They are intrinsically unstable and participate in chemical reactions
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A gas cylinder contains exactly 15 moles of oxygen gas (O2) how many molecules of oxygen are in a cylinder
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A sample of oxygen gas has a volume of 3.24 L at 29°C. What volume will it occupy at 104°C if the pressure and number of mol are
ohaa [14]

<u>Answer:</u> The final volume of the oxygen gas is 4.04 L

<u>Explanation:</u>

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=3.24L\\T_1=29^oC=(29+273)K=302K\\V_2=?\\T_2=104^oC=(104+273)K=377K

Putting values in above equation, we get:

\frac{3.24L}{302K}=\frac{V_2}{377K}\\\\V_2=4.04L

Hence, the final volume of the oxygen gas is 4.04 L

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2 years ago
the concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock
julsineya [31]

Answer:

5.0 x 10⁹ years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
  • The half-life of K-40 = 1.251 × 10⁹ years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (K-40) ([A₀] = 100%).

[A] is the remaining concentration of (K-40) ([A] = 6.25%).

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.

∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.

8 0
3 years ago
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