Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Answer:
D
Explanation:
Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa
Saying that the smaller the pKa value..the stronger the acid is.
Explanation:
from the equation 1 mole of O2 will give 2 moles of H2O then 6.0 moles of O2 will give x
6.0*2 moles/ 1 mole
= 12 moles
this implies that, 6.0 moles of O2 will give = 12 moles of water