C The number and types of bonds within the molecule.
Explanation:
In a molecule, the number and types of bonds present determines the amount of available energy therein.
When bonds are broken or formed, energy is usually released.
- Elements combine with one another in order to attain stability in this state.
- Through this process, they form bonds by attraction.
- Where atoms exchange their valence electrons by losing or gaining it, electrovalent bonds form.
- In covalent molecules, electrons are usually shared between atoms.
- An attraction result from this type of interaction.
- The bond formed stores energy in the process.
- When bonds are broken, energy is usually released. The energy accrues when the bond was being formed.
- In molecules, we have covalent bond.
Learn more:
Bond brainly.com/question/7213980
Covalent bonds brainly.com/question/5258547
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Answer:
The answer is C. Hope this helps you out!
<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
Answer:
molar mass = 180.833 g/mol
Explanation:
- mass sln = mass solute + mass solvent
∴ solute: unknown molecular (nonelectrolyte)
∴ solvent: water
∴ mass solute = 17.5 g
∴ mass solvent = 100.0 g = 0.1 Kg
⇒ mass sln = 117.5 g
freezing point:
∴ ΔTc = -1.8 °C
∴ Kc H2O = 1.86 °C.Kg/mol
∴ m: molality (mol solute/Kg solvent)
⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)
⇒ m = 0.9677 mol solute/Kg solvent
- molar mass (Mw) [=] g/mol
∴ mol solute = ( m )×(Kg solvent)
⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )
⇒ mol solute = 0.09677 mol
⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )
⇒ Mw solute = 180.833 g/mol
