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ryzh [129]
4 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]4 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]4 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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The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). If an object’s mas
Murljashka [212]
Weight=mass×g=10×9.8=98N
<span>hope it helped :-)</span>
6 0
3 years ago
What would be the mass in grams of 1.204 x 1024 molecules of sulfur dioxide
melomori [17]

Answer:

mass of sulfur = 96 g

Explanation:

no of moles of sulfur dioxide in 1.204\times 10^{24} molecules = \frac{1.204\times 10^{24}}{avagadro number }= \frac{1.204\times 10^{24}}{6.023\times 10^{23}}

                                       = 2 moles

therefore mass of sulfur dioxide = moles×atomic number

                                                      =2×(16+32)

                                                      =96

6 0
3 years ago
Which group would Ns2np2 go into
FrozenT [24]

Answer:

The group 15 elements: the pnicogens

Explanation:

The group 15 elements, nitrogen, phosphorus, arsenic, antimony and bismuth, all have the general valence shell electronic configuration ns2np3. They can all exist in the +3 or +5 oxidation state, with the +3 state increasing in stability as we move vertically down the group.

3 0
4 years ago
Read 2 more answers
For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
Question 13 of 32
jarptica [38.1K]

Answer:

A

Explanation:

this should be obvious if you read it haha

3 0
3 years ago
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