Answer:
80.0 g Na and 20.0 g N2.
Explanation:
This means the limiting reactant determines the maximum mass of the product formed.
B. At the equivalence point of a titration of the [H+] concentration is equal to 7.
<h3>What is equivalence point of a titration?</h3>
The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.
At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.
At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;
H⁺ + OH⁻ → H₂O
The pH of resulting solution is 7.0 (neutral).
Thus, the pH at the equivalence point for this titration will always be 7.0.
Learn more about equivalence point here: brainly.com/question/23502649
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We know that:
Molar Mass H2O: 18 g/mol
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>
<span>Using formula:
V= [mole fraction x molar mass] / density </span>
<span>mH20: 0.9947 * 18
= 17.9046 / 1 g/mL
= 17.9046 </span>
<span>morg: 0.0053 * 164
= 0.8692/ 1.05 g/mL
= 0.8278 </span>
<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>
Yotal volume = 30 mL; therefore,
<span>0.0442 = (volume eugenol/30) </span>
<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
= 1.44/1.05
= about 1.37 mL </span>
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
The average kinetic energy of 1 mole of a gas at -32 degrees Celsius is:
3.80 x 103 J
The relationship between volume and temperature of a gas, when pressure and moles of a gas are held constant, is: V*T = k.
FALSE
The relationship between moles and volume, when pressure and temperature of a gas are held constant, is: V/n = k. We could say then, that:
If the moles of gas are tripled, the volume must also triple.
If the temperature and volume of a gas are held constant, an increase in pressure would most likely be caused by an increase in the number of moles of gas.
TRUE
If the vapor pressure of a liquid is less than the atmospheric pressure, the liquid will not boil.
TRUE
35 - AB
36 - BD
33 - true
34 - False
20 - 6
21 - orthohombic
