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3 years ago

Describe how molecular motion and distance between molecules change as ice is heated to become a liquid and then a gas

Chemistry

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

Described in the explanation section

Explanation:

Ice is the solid phase of water and this phase, the water molecules usually have very little energy and therefore can't detach from each other. This is because the water molecules are closely packed together in a regular arrangement normally called lattice.

Now, when the ice is heated to become a liquid, the energy of the water molecules increases and thus some of them will have enough energy to overcome the intermolecular forces that had been keeping them closely packed in the ice phase. This will then make the water molecules to move further apart thereby forming liquid water.

Now, when the molecules are heated further, the liquid water will now turn into water vapor, which is the gas phase of water. The molecules in this gas phase will have more energy than in the liquid phase because the average distances between each of them will be much larger than the size of the molecules themselves.

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NNADVOKAT [17]

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3 years ago

The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

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Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

$t_{\frac{1}{2}} = \frac{0.693}{k}$

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

$ln(0.35) = -(1.325 *10^{-4})(t)$

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3 years ago

Which of the following is/are true regarding initial and maximal velocity of enzyme-mediated reactions?

kenny6666 [7]

Answer:

a. True

b. True

c. False

d. True

Explanation:

a). A a very low substrate concentration , $$S$ . Thus according to the Machaelis-Menten equation becomes

$V_0 = \frac{V_{max} \times [S]}{Km}$

Here since the $V_0$ varies directly to the substrate concentration [S], the initial velocity is lower than the maximal velocity. Thus option (a) is true.

b). The Michaelis -Menten kinetics equation states that :

$V_0 = \frac{V_{max} \times [S]}{Km+[S]}$

Here the initial velocity changes directly with the substrate concentration as $V_0$ is directly proportional to [S]. But $V_{max}$ is same for any particular concentration of the enzymes. Thus, option (b) is true.

c). As the substrate concentration increases, the initial velocity also increases. Thus option (c) is false.

d). Option (d) explains the procedures to estimate the initial velocity which is correct. Thus, option (d) is true.

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3 years ago