Response:
65.14 g
Clarification:
The entropy change resulting from mixing two metals is
![\delta S= R[(n_{cu}+n_{Ni})ln(n_{cu}+n_{Ni})-n_{cu}lnn_{cu}-n_{Ni}lnn_{Ni}]](https://tex.z-dn.net/?f=%5Cdelta%20S%3D%20R%5B%28n_%7Bcu%7D%2Bn_%7BNi%7D%29ln%28n_%7Bcu%7D%2Bn_%7BNi%7D%29-n_%7Bcu%7Dlnn_%7Bcu%7D-n_%7BNi%7Dlnn_%7BNi%7D%5D)
Here, R is the real gas constant, n_{cu} denotes the moles of copper, while n_{Ni} signifies the moles of nickel.
Therefore, determining the moles of Nickel yields

=1.7 moles
Then substituting n_{Ni} = 1.7 into the equation produces ΔS= 15,
with R= 8.314, and solving gives
![15= R[(n_{cu}+1.7)ln(n_{cu}+1.7)-n_{cu}lnn_{cu}-1.7ln1.7]](https://tex.z-dn.net/?f=15%3D%20R%5B%28n_%7Bcu%7D%2B1.7%29ln%28n_%7Bcu%7D%2B1.7%29-n_%7Bcu%7Dlnn_%7Bcu%7D-1.7ln1.7%5D)
Upon resolving, the quantity of moles of copper in the mixture equals
n_cu= 1.025
Thus, the copper mass is calculated as n_cu×M_cu = 1.025×63.55 = 65.14 g
Consequently, the required mass is found to be = 65.14 g
THE GREEN HOSE: Define the (x,y) coordinates at a height of 4 feet, which corresponds to where Majra holds the green hose. This indicates the equation for the green hose takes the form y = a(x - h)² + 4. Water from the hose lands on the ground 10 feet away from Majra, thus y(10) = -4. Given that the curve passes through (0,0), this leads to ah² + 4 = 0; therefore, ah² = -4. To satisfy the previous equation, we find a(10 - h)² + 4 = -4, simplifying to a(10 - h)² = -8. Dividing (3) by (4) gives a ratio of h²/(10-h)² = 1/2, leading to 2h² = (10 - h)² = 100 - 20h + h², and resolving yields h² + 20h - 100 = 0. Applying the quadratic formula, we get x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142. We discard the negative solution. The vertex locates at (4.142, 4). From (3), we deduce a = -4/4.142² = -0.2332, leading to the equation for the green hose: y = 0.2332(x - 4.142)² + 4. THE RED HOSE: The vertex of the red hose is positioned at (3,7), represented by the equation y = -(x-3)² + 7. A graph depicting y(x) for both hoses is included in the attached figure. Answers: a. The red hose throws water higher. b. The green hose's equation is y = -0.2332(x - 4.124)² + 4, starting at a height of 4 feet. c. The feasible domain for the green hose is between 0 ≤ x ≤ 10 feet, with the corresponding range being -4 ≤ y ≤ 4 feet.
The city evaluates the continuous increase of carbon monoxide from different origins each year. According to calculations, in the year "C: 2019"<span> (rounded to the closest whole number), the concentration of CO will surpass the allowed threshold.
If this is not correct, feel free to inform me and I will find out the right answer. However, I am confident this is accurate.:) </span>
The moment the body impacts the ground, two types of Forces are produced: the gravitational pull and the Normal Force. This aligns with Newton's third law, indicating that every action has an equal and opposite reaction. If the downward force of gravity is directed toward the earth, the reactionary force from the block acts upwards, equivalent to its weight:
F = mg
Where,
m = mass
g = gravitational acceleration
F = 5*9.8
F = 49N
Consequently, the answer is E.
Answer:
the temperature on the left side is 1.48 times greater than that on the right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
It is known that


n and v are constant on both sides. Therefore we have

..............1
let the final pressure be P and the temperature 

..................2
similarly
.............3
divide equation (2) by equation (3)
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, the left side temperature equals 1.48 times the right side temperature