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MissTica
3 years ago
6

Calculate the density of gold (in units of g/mL) if a 5.00 cm cube has a mass of 2.41kg

Chemistry
2 answers:
ikadub [295]3 years ago
6 0
Turn 2.41 kg into grams :

2.41 * 1000 => 2,410 g

1 cm³ = 1 mL , so:

5.00 cm³ = 5.00 mL

Therefore:

D = mass / volume

D = 2,410 / 5.00

D = 482 g/mL



levacccp [35]3 years ago
5 0
The volume of a 5cm cube is 0.5dm*0.5dm*0.5dm = 0.125dm^3
1dm^3 = 1l
So 0.125dm^3 = 125mL
This gives us 2410/125 = 19.28
So the density of gold is 19.28g/mL
Final check: Gold is indeed 19.3g/mL
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3 years ago
Draw a Lewis structure for C2H3Cl . Include all hydrogen atoms and show all unshared electron pairs. None of the atoms bears a f
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Answer:

See attached picture.

Explanation:

Hello!

In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.

In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.

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5 0
3 years ago
The formula K₂S indicates that _______.
Pachacha [2.7K]
The subscriot 2 means that in the formula there are two parts of K, and the subscript 1 (implicit) for S, indicates that there is one part of S.

This is, the formula gives the ratio of the elements K and S in the compound, which is:

2 atoms of K : 1 atom of S.

Answer: there are 2 atoms of K and 1 atom of S in a molecule of K2S.
5 0
3 years ago
3.0 x 108 m/s is the
Veseljchak [2.6K]

Answer:

Speed of light

Explanation:

The value 3.0 x 10⁸m/s is taken as the speed of light.

It is a constant.

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4 0
3 years ago
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PIT_PIT [208]
The calculation for such a question can be achieved via Avogadro hypothesis

We know molar mass of CO2 is 44g/mole which is the sum of atomic masses i.e; C and 2 oxygen atoms

Molar mass of CO2 =12(C)+2*16(O) = 44 g/mole will contain 6.023 ※10^23 CO2 molecules ..

44g/mole = 6.023 ※10^23 CO2 molecules

=> 1g = (6.023/44) ※10^23 CO2 molecules

==> 8.80g = 8.80(6.023÷44)10^23 = 1.2046 ※10^23 molecules of CO2….

Thus there r 1.2046 ※10^23 molecules of CO2 in 8.80g

if u need to calculate no. of carbon atoms then multiply result by 1 and if u need no of oxygen atoms in 8.80g of co2 then multiply the result by 2 ….
7 0
2 years ago
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