Answer:
0.897 J/g°C
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Mass (M) of substance = 155g
Initial temperature (T1) = 25.0°C
Final temperature (T2) = 40°C
Change is temperature (ΔT) = T2 – T1 = 40°C – 25.0°C = 15°C
Heat Absorbed (Q) = 2085 J
Specific heat capacity (C) of the substance =?
Step 2:
Determination of the specify heat capacity of the substance.
Applying the equation: Q = MCΔT, the specific heat capacity of the substance can be obtained as follow:
Q = MCΔT
C = Q/MΔT
C = 2085 / (155 x 15)
C = 0.897 J/g°C
Therefore, the specific heat capacity of the substance is 0.897 J/g°C
Answer:
the conjugate base of the compound is
C6h6O2/6-
Explanation:
subscript
Answer:
emulsion
Explanation:
An emulsion is a mixture of two or more liquids that are normally immiscible. Emulsions are part of a more general class of two-phase systems of matter called colloids.
Weather refers to short term atmospheric conditions while climate is the weather of a specific region averaged over a long period of time. Climate change refers to long-term changes.
Answer:
Explanation:
Density is m/V. Also, 1 liter = 1000 
. So, we get 0.890/(5*1000) = 
g/cm^3. You can convert this to kg/m^3 as well by multiplying it by 10. Depends which one you want.
