Answer: Time needed: 2.5 s
Distance covered: 31.3 m
Explanation:
I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by
on one side of the equation and solve by plugging your values
To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation
Explanation:
Answer:
Protons, electrons, same in that order is the answer.
Explanation:
this has to do with the periodic table information.
Im sure that the answer is C, correct me if im wrong.
Answer: The heat of reaction (ΔHrxn) for the reaction is -164.9kJ
Explanation:
The given balanced chemical reaction is,

To calculate the enthalpy of reaction
.

![\Delta H^o=[n_{CaCl_2}\times \Delta H_f^0_{(CaCl_2)}+n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{CaCO_3}\times \Delta H_f^0_{(CaCO_3)+n_{HCl}\times \Delta H_f^0_{(HCl)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BCaCl_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CaCl_2%29%7D%2Bn_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CO_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BCaCO_3%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28CaCO_3%29%2Bn_%7BHCl%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28HCl%29%7D%5D)
where,

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times -877.1)+(1\times -393.51)+(1\times -285.8)]-[(1\times -1206.9)+(2\times -92.30)]=-164.9kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20-877.1%29%2B%281%5Ctimes%20-393.51%29%2B%281%5Ctimes%20-285.8%29%5D-%5B%281%5Ctimes%20-1206.9%29%2B%282%5Ctimes%20-92.30%29%5D%3D-164.9kJ)
Therefore the heat of reaction (ΔHrxn) for the reaction is -164.9kJ
Percent composition of a molecule is found through the molecular formula (Mass of each element found in a single mole of the molecule ÷ total molar mass of the molecule). This tells you how much of a specific element makes up the molecule.
Also be aware that 2 molecules can have the same % composition but different formulas.