Answer:
ΔE negative,ΔE positive,ΔE positive
Explanation:
Internal energy change of a system is positive when the system acquires energy from outside and internal energy change is negative when energy is being liberated from the system.
a Sweat evaporates from skin,cooling the skin ΔE is negative as heat energy is liberated through evaporation of sweat.
b A balloon expand against an external pressure ΔE positive because external energy is required to expand the balloon.
c An aqueous chemical reaction mixture is warmed with an external flame ΔE is positive because heat energy is provided by the external flame to warm the aqueous chemical reaction.
<u>Answer:</u> The 
for the reaction is -521.6 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of fluorine and water follows:
The intermediate balanced chemical reaction are:
(1) 
( × 2)
(2) 
The expression for enthalpy of reaction follows:
![\Delta H^o_{rxn}=[2\times \Delta H_1]+[1\times (-\Delta H_2)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B2%5Ctimes%20%5CDelta%20H_1%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_2%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(2\times (-546.6))+(1\times (571.6))]=-521.6kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-546.6%29%29%2B%281%5Ctimes%20%28571.6%29%29%5D%3D-521.6kJ)
Hence, the for the reaction is -521.6 kJ.
Answer:
m H2O = 56 g
Explanation:
∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:
⇒ - (mCΔT)Al = (mCΔT)H2O
∴ m Al = 25.0 g
∴ Mw Al = 26.981 g/mol
⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al
⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C
⇒ Q Al = 1327.64 J
∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C
⇒ mH2O = 55.722 g ≅ 56 g
Answer:
C. 10/9 gallons
Explanation:
Given that
V= 20 gallons
Volume of ethanol = 5% of V = 0.05 x 20 = 1 gallons
Volume of the gasoline = 95 % of V = 19 gallons
Lets take x gallons of ethanol is added to achieve optimum performance.
(1+x)/(20+x) = 10 %
(1+x)/(20+x) =0.1
1+ x = 0.1 ( 20+ x)
1+ x = 2 + 0.1 x
1 = 0.9 x
x= 10 /9 gallons
So the option C is correct.
C. 10/9 gallons
