Formation of mountains by one continental plate moving into another (Indian plate pushing north, forming the Himalayas)
Vocanoes from an oceanic plate being submerged beneath another plate and melting, causing liquid hot magma to rise to the surface
A CH compound is combusted to produce CO2 and H2O
CnHm + O2 -----> CO2 + H2O
Mass of CO2 = 23.1g
Mass of H2O = 10.6g
Calculate by mass of the compounds
For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
Calculate the moles for C and H
6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
Divides by both mole entities with smallest
C = 0.524 / 0.524 = 1 x 4 = 4
H = 1.17 / 0.524 = 2.23 x 4 = 10
The empirical formula is C4H10.
Answer
The empirical formula is CrO₂Cl₂
Explanation:
Empirical formula is the simplest whole number ratio of an atom present in a compound.
The compound contain, Chromium=33.6%
Chlorine=45.8%
Oxygen=20.6%
And the molar mass of Chromium(Cr)=51.996 g mol.
Chlorine containing molar mass (Cl)= 35.45 g mol.
Oxygen containing molar mass (O)=15.999 g mol.
Step-1
Then,we will get,
Cr=
mol
Cl= 
mol.
O=
mol.
Step-2
Divide the mole value with the smallest number of mole, we will get,
Cr=
Cl=
O=
Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)
Answer:
8.37 grams
Explanation:
The balanced chemical equation is:
C₆H₁₂O₆ ⇒ 2 C₂H₅OH (l) + 2 CO₂ (g)
Now we are asked to calculate the mass of glucose required to produce 2.25 L CO₂ at 1atm and 295 K.
From the ideal gas law we can determine the number of moles that the 2.25 L represent.
From there we will use the stoichiometry of the reaction to determine the moles of glucose which knowing the molar mass can be converted to mass.
PV = nRT ⇒ n = PV/RT
n= 1 atm x 2.25 L / ( 0.08205 Latm/kmol x 295 K ) =0.093 mol CO₂
Moles glucose required:
0.093 mol CO₂ x ( 1 mol C₆H₁₂O₆ / 2 mol CO₂ ) = 0.046 mol C₆H₁₂O₆
The molar mass of glucose is 180.16 g/mol, then the mass required is
0.046 mol x 180.16 g/mol = 8.37 g
Answer:
both will be at liquid state. the particles will move rapidly in all directions and will collide with other particles in random motion
