Depends on the grade of the test. But for the most part if it’s below a 78, it’ll drop, If it’s above it’ll raise. It shouldn’t drop or raise it more than 3 point/percent.
Answer:
ΔE = 150 J
Explanation:
From first law of thermodynamics, we know that;
ΔE = q + w
Where;
ΔE is change in internal energy
q is total amount of heat energy going in or coming out
w is total amount of work expended or received
From the question, the system receives 575 J of heat. Thus, q = +575 J
Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.
Thus;
ΔE = 575 + (-425)
ΔE = 575 - 425
ΔE = 150 J
Answer: 15062.4 Joules
Explanation:
The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of food = 200.0g
C = 4.184 j/g°C
Φ = (Final temperature - Initial temperature)
= 83.0°C - 65.0°C = 18°C
Then, Q = MCΦ
Q = 200.0g x 4.184 j/g°C x 18°C
Q = 15062.4 J
Thus, 15062.4 joules of heat energy was contained in the food.
Answer:
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
Explanation:
A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.
A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.
The equations for the neutralizations that occurred upon addition of HCl or NaOH are;
HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)
NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)
