Answer:
- According to the law <br> Mass of reactants = mass of product, here <br> `underset(10 g)(CaCO_(3))rarr underset(4.4 g)(CO_(2))+underset(x)(CaO)` <br> Hence, x = 10 g - 4.4 g = 5.6 g <br> Which is mass of CaO.d
- In the first compound <br> Hydrogen = 5.93 % <br> Oxygen = `(100-5.93)% = 94.07 %` <br> In the second compound <br> Hydrogen = 11.2 % <br> Oxygen `= (100-11.2)%=88.8%` <br> In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen `=(94.07)/(5.93)=15.86` parts ...
- (The ratio of Cu combining with fixed weight of oxygen in black and red oxide is 1 : 2 respectively. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.) {Check something more in the above attachment!}
- Refer to the above attachment
Explanation:
<em>Mark </em><em>this </em><em>answer </em><em>as </em><em>brainlest </em><em>answer</em><em>!</em><em>!</em>
Answer:
I dont know at all and that is confusing.
Hydrogen sulfide has a low boiling point because it is only a slightly polar molecule, meaning that it only has weak dipole-dipole intermolecular forces. Because of its relatively weak intermolecular forces, it has a low boiling point.
Answer : The thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole
Explanation :
First we have to calculate the heat released by the combustion.

where,
q = heat released = ?
= specific heat of calorimeter = 
= change in temperature = 
Now put all the given values in the above formula, we get:


Thus, the heat released by the combustion = 70.43 kJ
Now we have to calculate the molar enthalpy combustion.

where,
= molar enthalpy combustion = ?
q = heat released = 70.43 kJ
n = number of moles cymene = 

Therefore, the thermal energy produced during the complete combustion of one mole of cymene is -7193 kJ/mole