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g100num [7]
4 years ago
5

Describe the difference between a ball-and-stick model and a space-filling model of a compound.

Chemistry
1 answer:
Lisa [10]4 years ago
4 0
A space-filling model shows the relative amount of space each atom takes up. In other words, a space-filling model can show relative sizes of atoms. However, unlike ball-and-stick or structural models, space-filling models do not show bond lengths clearly. Bonds are not really like sticks in a ball-and-stick model.
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Explain in terms of intermolecular forces why water has an unusually high heat of fusion?
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Explanation:

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A malfunction at a power station causes a power surge in certain homes near the station. Some televisions and computers begin ma
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Sandstone is a sedimentary rock which of the following characteristics would you look for in a piece of sand stone
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3 years ago
The following equations are half reactions and reduction potentials. Ag+ (aq) + e– Ag(s) has a reduction potential of +0.80 V. C
Lady bird [3.3K]

Answer : The correct option is, (A) silver ion gains electrons more easily and is a stronger oxidizing agent than a chromium(III) ion.

Explanation :

The given half reaction are :

1st half reaction : Ag^+(aq)+e^-\rightarrow Ag(s)

The reduction potential of this reaction = +0.80 V

2nd half reaction : Cr^{3}+(aq)+3e^-\rightarrow Cr(s)

The reduction potential of this reaction = -0.74 V

From the reduction potentials, we conclude that the reaction which have positive reduction potential, they will gain electrons more easily and reduced itself and act as a stronger oxidizing agent.

Or we can say that the reaction which have negative reduction potential, they will lose electrons more easily and oxidized itself and act as a stronger reducing agent.

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5 0
4 years ago
Read 2 more answers
Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years
Sliva [168]

Answer : The time passed in years is 2.83\times 10^3\text{ years}

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

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

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = let initial amount of the reactant  = X g

a - x = amount left after decay process = 71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}

t=2831.00\text{ years}=2.83\times 10^3\text{ years}

Therefore, the time passed in years is 2.83\times 10^3\text{ years}

8 0
3 years ago
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