Answer:
Explanation:
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In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:
Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:
Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:
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We have that the molecular weight (3sf) of the compound (g/mol)
From the question we are told
A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).
Generally the equation for the Rouault's law is mathematically given as
P=P_0 N
Therefore
The molecular weight (3sf) of the compound (g/mol)
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A. It can be as important to a story as a character
Answer: The heat required is 6.88 kJ.
Explanation:
The conversions involved in this process are :
Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of ethanol = 25.0 g
= specific heat of solid ethanol= 0.97 J/gK
= specific heat of liquid ethanol = 2.31 J/gK
n = number of moles of ethanol = 
= enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole
= change in temperature
The value of change in temperature always same in Kelvin and degree Celsius.
Now put all the given values in the above expression, we get
![\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B25.0%20g%5Ctimes%200.97J%2FgK%5Ctimes%20%28-114-%28-135%29K%5D%2B0.534mole%5Ctimes%205020J%2Fmole%2B%5B25.0g%5Ctimes%202.31J%2FgK%5Ctimes%20%28-50-%28-114%29%29K%5D)
(1 KJ = 1000 J)
Therefore, the heat required is 6.88 kJ
