Answer:
0.8 mL of protein solution, 9.2 mL of water
Explanation:
The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:
C₁V₁ = C₂V₂
We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.
V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL
Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:
(10 mL - 0.8 mL) = 9.2 mL
Answer : The value of 
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction 
.
![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)
Now we have to calculate the Gibbs free energy of reaction 
.
As we know that,
At room temperature, the temperature is 500 K.
Therefore, the value of for the reaction is -959.1 kJ
Answer:
MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2
Explanation:
I'm assuming you want to balance it so...
The first thing I see is that there are two chlorines on the reactant side and one on the product side
Adding a coefficient of 2 would get 2AgCl2
Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.
If that is not what you are looking for let me know!
The kinetic energy of the shopping cart increases. This speeds up the energy rate of friction between the wheel and the floor giving off more kinetic energy so the speed rate can increase (depending on how much your pushing it) Hope this helps :)
Answer:
The interaction reaction of calcium oxide and water leads to the formation of calcium hydroxide, this reaction is exothermic, which indicates that energy is released.
CaO(s) + H2O ⇒ Ca(OH)2
Explanation:
Calcium oxide reacts violently with water, causing it to reach 90 C. It then forms calcium hydroxide, also called slaked lime. It is heated because it releases energy in the form of heat, because the new substance formed (slaked lime) has less internal energy than the original and this is released in the form of heat. As a general rule, acid and base solutions are exothermic.
