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Anestetic [448]
3 years ago
6

What is the law of conservation of mass

Chemistry
2 answers:
e-lub [12.9K]3 years ago
6 0
Mass can only be transferred
OleMash [197]3 years ago
4 0
Mass is not created nor can it be destroyed, similar to the Law of Conservation of Matter.
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What do you use to determine molar mass?
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the molar mass of any element can be determined by finding the atomic mass of the element on the periodic table for example, if the atomic mass of sulfer is 32.066 amu, then it's molar mass is 32.066 g / mol

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For the gaseous reaction of carbon monoxide and chlorine to form phosgene (COCl₂):(b) Assuming that ΔS° and ΔH° change little wi
maksim [4K]

ΔG° at 450. K is -198.86kJ/mol

The following is the relationship between  ΔG°,  ΔH, and  ΔS°:

ΔH-T ΔS = ΔG

where  ΔG represents the common Gibbs free energy.

the enthalpy change,  ΔH

The temperature in kelvin is T.

Entropy change is  ΔS.

ΔG° = -206 kJ/mol

ΔH° equals -220 kJ/mol

T = 298 K

Using the formula, we obtain:

-220kJ/mol -T ΔS° = -206kJ/mol

220 kJ/mol +206 kJ/mol =T ΔS°.

-T ΔS = 14 kJ/mol

for ΔS-14/298

ΔS=0.047 kJ/mol.K

450K for the temperature Completing a formula with values

ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol

ΔG° = -220 kJ/mol + 21.14 kJ/mol.

ΔG°=198.86 kJ/mol

Learn more about ΔG° here:

brainly.com/question/17214066

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7 0
2 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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