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ryzh [129]
3 years ago
11

A 13.0 kg iron weightlifting plate has a volume of 1650 cm3 . What is the density of the iron plate in g/cm3?

Chemistry
2 answers:
Salsk061 [2.6K]3 years ago
8 0
11.0 kg = (11.0 kg)(1000 g/kg) = 11000 g
(11000 g)/(1400 cm3) = 7.857 g/cm3
Simplified = 7.86 g/cm3
Oksi-84 [34.3K]3 years ago
5 0

Explanation:

It is known that density is the amount of mass divided by volume.

Mathematically,     Density = \frac{mass}{volume}

It is given that mass is 13.0 kg or 13000 g (as 1 kg = 1000 g). And, volume is 1650 cm^{3}.

Therefore, calculate the density as follows.

                    Density = \frac{mass}{volume}

                                 = \frac{13000 g}{1650 cm^{3}}    

                                 = 7.87 g/cm^{3}

Thus, we can conclude that density of the given iron plate is 7.87 g/cm^{3}.

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Answer:

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Explanation:

The shape and the bond orientation of molecules and ions are both explained by the valences shell electron pair repulsion theory (VSEPR).

Ammonia, NH_3, is a molecule which contains three N-H bonds, as well as one lone pair on nitrogen. According to the VSEPR theory, molecules try to acquire a shape which would minimize the repulsion exhibited by the electron clouds present, that is, between the bonding (shared in a bond) and non-bonding (lone pair) electrons.

In VSEPR, our main step is to calculate the steric number, this is the sum of the number of bonds (ignoring the multiplicity of any bond) and the lone pairs on a central atom. In ammonia, we have 3 bonds and 1 lone pair, totaling to a steric number of 4. A steric number of 4 without any lone pairs on a central atom and just bonds would yield a tetrahedral shape with bond angles of 109.5^o.

Now, in this case, since we have a lone pair instead of a bond, it is repelling stronger decreasing the bond angles to about 107^o.

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