Answer:
Normalidad = 4N
%p/V = 27.6%
Explanation:
La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:
2moles * (2eq/mol) = 4eq / 1L = 4N
El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:
%p/V:
Masa K2CO3 -Masa molar: 138.205g/mol-
2moles * (138.205g/mol) = 276g K2CO3
Volumen:
1L * (1000mL/1L) = 1000mL
%p/V:
276g K2CO3 / 1000mL * 100
<h3>%p/V = 27.6%</h3>
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>
Answer:
Project 3.
Explanation:
Project 3's anticipated cost is 12 to 17 million dollars. It is a <em>lower </em>anticipated cost than Project 2 and Project 4, but <em>higher</em> than Project 1 by one million dollars at maximum cost anticipation. Additionally, the percentage of wildlife to benefit is 70-80%, which is <em>second</em> to the most wildlife to benefit which is Project 4 at 75-80%.
And finally, for community support for Project 3 - the chart lists it as high. This outclasses Project 2 and Project 4, but balances with Project 1. However, Project 1 costs 13 to 16 million dollars and <em>only</em> benefits 15-25% of wildlife.
Answer: C2HNO3
Explanation: C = 27.59/12.011 = 2.297
H = 1.15/1.008 = 1.1409
N = 16.09/14.007 = 1.1487
O = 55.17/15.999 = 3.4483
Divide by smallest result:
C = 2
H=1
N=1
O = 3
Empirical formula = C2HNO3
