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satela [25.4K]
3 years ago
6

How many cm3 are contained in 3.83 x 107 mm3?

Chemistry
1 answer:
gulaghasi [49]3 years ago
5 0

Answer:

3.38 x 10⁴cm³

Explanation:

Given that:

      Volume of the substance  = 3.83 x 10⁷mm³

The problem entails use to convert from mm³ to cm³;

                    1000mm³   = 1cm³

So;

          3.83 x 10⁷mm³ will produce \frac{ 3.83 x 10^{7} }{1000}    = 3.38 x 10⁴cm³

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What is the molecular geometry if you have 3 single bonds and 1 lone pair around the central atom?
Elodia [21]

Answer:

There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.

Explanation:

There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.

7 0
3 years ago
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0
3 years ago
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D) kinetinc to electric
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Answer:

1.6

2.8

3.6

Explanation:

4 0
3 years ago
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Devices that mix air or oxygen with flammable gasses cannot be used unless approved by
Feliz [49]

Flammable gases are gases that have the tendency to <u>explode (burst into flames)</u> when they come in contact with the <u>appropriate quantities of air, oxygen, or any suitable oxidant.</u>

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Devices that mix air or oxygen with flammable gasses cannot be used unless approved by an <u>authorized or approved personnel.</u>

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  • Examples of flammable gases are hydrocarbons such as <u>Propane, Acetylene</u>, e.t.c.

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<u></u>

<u></u>

To learn more, visit the link below:

brainly.com/question/3702349

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2 years ago
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