Convalent Bond energy is low
Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
Answer:
This atomic model has changed over time. Scientists used the model to make predictions. Sometimes the results of their experiments were a surprise and they did not fit with the existing model. Scientists changed the model so that it could explain the new evidence.
Explanation:
Answer: -227 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)

Therefore, the enthalpy change for
is -227 kJ.