Answer:
![\Delta _{comb}H=-2,265\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_%7Bcomb%7DH%3D-2%2C265%5Cfrac%7BkJ%7D%7Bmol%7D)
Explanation:
Hello!
In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:
![Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}](https://tex.z-dn.net/?f=Q%20_%7Bcomb%7D%3D-C_%7Bcalorimeter%7D%5CDelta%20T_%7Bcalorimeter%7D)
In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:
![Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J](https://tex.z-dn.net/?f=Q%20_%7Bcomb%7D%3D-2980%5Cfrac%7BJ%7D%7B%5C%C2%B0C%7D%20%2A11.95%5C%C2%B0C%5C%5C%5C%5CQ%20_%7Bcomb%7D%3D-35%2C611J)
Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:
![n=1.400g*\frac{1mol}{89.05g} =0.01572mol](https://tex.z-dn.net/?f=n%3D1.400g%2A%5Cfrac%7B1mol%7D%7B89.05g%7D%20%3D0.01572mol)
Thus, we obtain:
![\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol} \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J} \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_%7Bcomb%7DH%3D%5Cfrac%7BQ_%7Bcomb%7D%7D%7Bn%7D%3D%5Cfrac%7B-35%2C611J%7D%7B0.01572mol%7D%20%20%5C%5C%5C%5C%5CDelta%20_%7Bcomb%7DH%3D-2%2C265%2C331%5Cfrac%7BJ%7D%7Bmol%7D%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%20%5C%5C%5C%5C%5CDelta%20_%7Bcomb%7DH%3D-2%2C265%5Cfrac%7BkJ%7D%7Bmol%7D)
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Answer:
An acid
Explanation:
According to the Arrhenius theory of acids and bases "an Arrhenius acid is a compound that increases the concentration of H+ ions that are present when added to water"
Answer:
0.081g
Explanation:
first step:calculate the molar conc of the stock solution
molar conc=mass/molar mass ×1000/vol(ml)
molar mass of Co(N03)2=182.943
molar conc=3g/182.943 ×1000/100ml
=0.163M
To calculate the conc of the resulting solution we apply the dilution principle
n=cv
C1V1=C2V2
C1=conc of stock solution=0.163M
V1=4mL
V2=275ml(final volume)
by making 'C2" the subject C2=0.163×4/275
=0.0024M
by applying the formular for calculating molar conc of a solution,we can calculate the mass of CO(NO3)2 in the reulting solution
molar conc=mass/molar mass ×1000(L)/vol(mL)
mass=0.0024×182.943×275/1000
=0.1193g
Mass of CO(NO3)2 in the resulting solution=0.1199g
since CO+(NO3)2=CO(NO3)2
Percentage by mass of (NO3)2 in CO(NO3)2=(NO3)2/CO(NO3)2
=124/182.943×100
=67.8%
it means 67.8% of 0.1193g is the mass of (NO3)2=0.678×0.1199
=0.081g
Answer:
The molar mass and molecular weight of 3CaCl2 is 332.952.