Answer:
Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. ... Certain substances are soluble in all proportions with a given solvent, such as ethanol in water.
Explanation:
Don’t eat or drink in labs
Dress for the lab; don’t wear open toed shoes
Dispose of lab waste properly
No horse play
Don’t taste or sniff things in the lab
Tire your hair back
Answer:
1.69.
Explanation:
- The solution = 12.0 / 7.11 = 1.687 = 1.69.
- The rule of significant figures for division states that: the results are reported to the fewest significant figures.
- 12.0 contains 3 significant figures.
- 7.11 contains 3 significant figures.
So, the solution should contain 3 significant figures.
- Now, the issue id of rounding; In a series of calculations, carry the extra digits through to the final result, then round.
- If the digit to be removed is equal to or greater than 5, the preceding digit is increased by 1.
- The digit that should be removed is 7 that is larger than 5 so increase the preceding digit by 1.
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
