Rocks, rocks can be used for many things depending on which rock you are talking about some rocks are used for construction of cement mixtures. Granite and marble are used for counter tops and desks depending on what district you go to school in and how funded the schools are there. pumice is a rock found in waters around volcanoes and is used to smooth organic surfaces such as feet. miscellaneous rocks are also used to prevent erosion. non living things can also include carbon which is the bases of life itself. Carbon is an organic element which is used in the synthesis medicines such as aspirin and poisons such as dichloromethane which is not used to poison others but is toxic to humans, it is mainly used for different isolation of unknown substances. Depending on what R grouping is attached to the carbon can be the deciding factor of how it can be used and how it can react with the environment. Elements like helium can be used in gas chromatograms and put into balloons
Answer:
D number of protons or atomic mass
1) Chemical equation
<span>2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l)
2) Stoichiometric ratios
2 mol NH4Cl(s) : 54.8 KJ
3) Convert 24.7 g of NH4Cl into number of moles, using the molar mass
molar mass of NH4Cl = 14 g/mol + 4*1 g/mol + 35.5 g/mol = 53.5 g/mol
number of moles = mass in grams / molar mass
number of moles = 24.7 g / 53.5 g/mol = 0.462 moles
4) Use proportions:
2 moles NH4Cl / 54.8 kJ = 0.462 moles / x
=> x = 0.462 moles * 54.8 kJ / 2 moles = 12.7 kJ
Answer: 12.7 kJ
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Answer : The molecular formula of a compound is, 
Solution : Given,
Mass of C = 64.03 g
Mass of H = 4.48 g
Mass of Cl = 31.49 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of Cl = 35.5 g/mole
Step 1 : convert given masses into moles.
Moles of C = 
Moles of H = 
Moles of Cl = 
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = 
For H = 
For Cl = 
The ratio of C : H : Cl = 6 : 5 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula =
The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
Molecular formula = 
Therefore, the molecular of the compound is,
