Answer:
1.23 j/g. °C
Explanation:
Given data:
Mass of metal = 35.0 g
Initial temperature = 21 °C
Final temperature = 52°C
Amount of heat absorbed = 320 cal (320 ×4.184 = 1338.88 j)
Specific heat capacity of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 52°C - 21 °C
ΔT = 31°C
1338.88 j= 35 g ×c× 31°C
1338.88 j= 1085 g.°C ×c
1338.88 j/1085 g.°C = c
1.23 j/g. °C = c
This is pretty easy lol.... AS and AR
Answer: A. Exothermic reaction
Explanation: Enthalpy change for a reaction is sum of enthalpy of formation of products minus sum of enthalpy of formation of reactants.
When the energy level of reactants is above as compared to the products, the reaction is exothermic and when its opposite then reaction is endothermic.
From given information, the potential energy diagram starts at 380 kJ means the energy level of reactants is 380 kJ. It ends at 100 kJ means the energy of products is 100 kJ.
Enthalpy of reaction = 100 kJ - 380 kJ
Enthalpy of reaction = -280 kJ
Negative sign of enthalpy change indicates an Exothermic reaction.
Answer:
hope i helped!
pls mark brainliest if i am correct
3.07g H2
27.4/26.98/2x3x1.01x2=3.07
