I think it would be Bromine and Mercury, hope that helps
Answer:
2.79 °C/m
Explanation:
When a nonvolatile solute is dissolved in a pure solvent, the boiling point of the solvent increases. This property is called ebullioscopy. The temperature change (ΔT) can be calculated by:
ΔT = Kb*W*i
Where Kb is the ebullioscopy constant for the solvent, W is the molality and i is the van't Hoff factor.
W = m1/(M1*m2)
Where m1 is the mass of the solute (in g), M1 is the molar mass of the solute, and m2 is the mass of the solvent (in kg).
The van't Hoff factor represents the dissociation of the elements. For an organic molecule, we can approximate i = 1. Thus:
m1 = 2.00 g
M1 = 147 g/mol
m2 = 0.0225 kg
W = 2/(147*0.0225)
W = 0.6047 mol/kg
(82.39 - 80.70) = Kb*0.6047*1
0.6047Kb = 1.69
Kb = 2.79 °C/m
He answer in here is b.7.5 cubic meters. We know this when we do the following calculation:
<span>(10 m^3) x (1 / 1.5) x ((32 + 273) / (273)) = 7.5 m^3
</span>
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
Hey There!
To solve this question remember that 1 mole of ammonia is equal to 17G of ammonia.
Using that, we first multiply 17 by 7 = 24
Then we multiply 7 by .27 = 1.89
Finally, taking those and adding them together gives you the answer...
<u>25.89 grams :)</u>
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<u>Have a great day, I hope I helped.</u>
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<u>+ Brainliest & Five stars are always appreciated </u>