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dexar [7]
2 years ago
12

The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 x 10-8 cm. What is this distance in inche

s?
Chemistry
1 answer:
Sergio [31]2 years ago
5 0

Answer:

d=4.75\times 10^{-9}\ \text{inches}

Explanation:

Given that,

The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21\times 10^{-8}\ cm.

We need to convert this distance in inches.

We know that,

1 cm = 0.393 inches

We can solve it as follows :

1.21\times 10^{-8}\ cm=0.393\times 1.21\times 10^{-8}\\\\=4.75\times 10^{-9}\ \text{inches}

So, the distance between the centers of the two oxygen atoms is 4.75\times 10^{-9}\ \text{inches}.

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Lead (Pb) is used to make a number of different alloys. What is the mass of lead present in an alloy containing 0.15 mol of lead
Vesna [10]
Lead is 207.2 g/mol.
0.15 mol (\frac{207.2g}{mol}) = 31.08 g Pb
3 0
2 years ago
Which statement correctly compares the number of chromosomes in a body cell to that in a sex cell of humans?
ad-work [718]
The answer is C: "<span>A body cell has 46 chromosomes; a sex cell has 23"</span>
6 0
3 years ago
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Pls help
Dmitry_Shevchenko [17]

Answer:

The answer would be beryllium

You were probably confused but the correct answer would actually be a beryllium ion.

4 0
2 years ago
The half-life of radioactive substance is 2.5 minutes. what fraction of the origional radioactive remains after 10 mins
saul85 [17]
The answer is 1/16.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. (1/2)^{n} = x,
where:
<span>n - a number of half-lives
</span>x - a remained fraction of a sample

2. t_{1/2} = \frac{t}{n}
where:
<span>t_{1/2} - half-life
</span>t - <span>total time elapsed
</span><span>n - a number of half-lives
</span>
So, we know:
t = 10 min
<span>t_{1/2} = 2.5 min

We need:
n = ?
x = ?
</span>
We could first use the second equation to calculate n:
<span>If:
t_{1/2} = \frac{t}{n},
</span>Then: 
n = \frac{t}{ t_{1/2} }
⇒ n = \frac{10 min}{2.5 min}
⇒ n=4<span>
</span>
Now we can use the first equation to calculate the remained fraction of the sample.
<span>(1/2)^{n} = x
</span>⇒ x=(1/2)^4
<span>⇒x= \frac{1}{16}</span>
3 0
3 years ago
For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

3 0
3 years ago
Read 2 more answers
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