Answer:
Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.
Step-by-step explanation:
Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.
Answer:
C
Step-by-step explanation:
C is the right answer that you're looking for because the hundredth value in this number is 3 in 345 and the closest hundredth there is 300 so the number closest to hundredths is 2,684,300
Answer: ![x=-2.8](https://tex.z-dn.net/?f=x%3D-2.8)
Step-by-step explanation:
Given the following equation:
![0.7x+6=0.3(4x+6)-2x](https://tex.z-dn.net/?f=0.7x%2B6%3D0.3%284x%2B6%29-2x)
You can follow these steps in order to solve for "x" and find its value:
1. Apply Distributive property on the right side of the equation:
![0.7x+6=0.3(4x+6)-2x\\\\0.7x+6=1.2x+1.8-2x](https://tex.z-dn.net/?f=0.7x%2B6%3D0.3%284x%2B6%29-2x%5C%5C%5C%5C0.7x%2B6%3D1.2x%2B1.8-2x)
2. Subract
from both sides and add the like terms:
![0.7x+6-(0.7x)=1.2x+1.8-2x-(0.7x)\\\\6=-1.5x+1.8](https://tex.z-dn.net/?f=0.7x%2B6-%280.7x%29%3D1.2x%2B1.8-2x-%280.7x%29%5C%5C%5C%5C6%3D-1.5x%2B1.8)
3. Subtract 1.8 from both sides:
![6-(1.8)=-1.5x+1.8-(1.8)\\\\4.2=-1.5x](https://tex.z-dn.net/?f=6-%281.8%29%3D-1.5x%2B1.8-%281.8%29%5C%5C%5C%5C4.2%3D-1.5x)
4. Finally, divide both side of the equation by -1.5:
![\frac{4.2}{-1.5}=\frac{-1.5x}{-1.5}\\\\x=-2.8](https://tex.z-dn.net/?f=%5Cfrac%7B4.2%7D%7B-1.5%7D%3D%5Cfrac%7B-1.5x%7D%7B-1.5%7D%5C%5C%5C%5Cx%3D-2.8)
The answer is 38 since range=high - low = 58-20
Answer:
https://youtu.be/ssE-HVO4SY4