The specific heat, c is 0.75 J/g°C
<u>Explanation:</u>
Heat or Energy, Q = 1500J
Mass, m = 50g
T1 = 0°C
T2 = 40°C
Specific Heat, c = ?
We know,
Q = mcΔT
Q = mc(T2-T1)
1500 = 50 X c X (40-0)
1500 = 50 X c X 40
c = 1500/ 2000
c = 0.75 J/g°C
Therefore, the specific heat, c is 0.75 J/g°C
Answer:
131 K
Explanation:
The only variables are volume and temperature, so we can use Charles' Law:
V1/T1 = V2/T2
Data:
V1 = 2.00 L; T1 = 210 K
V2 = 1.25 L; T2 = ?
Calculation:
2.00/210 = 1.25/T2
Multiply each side by the lowest common denominator (210T2)
2.00T2 = 1.25 ×210
2.00T2 = 262.5
T2 = 262.5/2.00 = 131 K
The gas was cooled to 131 K.
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#4 (Recycling synthetic polymers is costly)
Rain bc it’s hot then cools
Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)