Answer:
i have an answer but i can only show you because my teacher helped my on it and wrote it down for me to remember! hope this helps!!!
Explanation:
A 1.00 g sample ofn-hexane (C6H14) under-goes complete combustion with excess O2ina bomb calorimeter. The temperature of the1502 g of water surrounding the bomb risesfrom 22.64◦C to 29.30◦C. The heat capacityof the hardware component of the calorimeter(everything that is not water) is 4042 J/◦C.What is ΔUfor the combustion ofn-C6H14?One mole ofn-C6H14is 86.1 g.The specificheat of water is 4.184 J/g·◦C.1.-9.96×103kJ/mol2.-7.40×104kJ/mol3.-1.15×104kJ/mol4.-4.52×103kJ/mol5.-5.92×103kJ/molcorrectExplanation:mC6H8= 1.00 gmwater= 1502 gSH = 4.184 J/g·◦CHC = 4042 J/◦CΔT= 29.30◦C-22.64◦C = 6.66◦CThe increase in the water temperature is29.30◦C-22.64◦C = 6.66◦C. The amount ofheat responsible for this increase in tempera-ture for 1502 g of water isq= (6.66◦C)parenleftbigg4.184Jg·◦Cparenrightbigg(1502 g)= 41854 J = 41.85 kJThe amount of heat responsible for the warm-ing of the calorimeter isq= (6.66◦C)(4042 J/◦C)= 26920 J = 26.92 kJ
Answer:First calculate the moles of selenium by multiplying the mass x molar mass 90.2 × 78.96 = 7122.2 moles. Number of moles multiplied by ...
Explanation:
Explanation:
The given data is as follows.
Density of vinegar = 1.0 g/ml
Specific heat capacity = 4.25 
= 
, and 
= 
Relation between enthalpy and specific heat is as follows.
Hence, putting the values into the above formula as follows.
= 
(as density = 
)
= - 315 J
Thus, we can conclude that the enthalpy of reaction is -315 J.
As the value is negative so, it means that heat is releasing. Hence, the reaction is exothermic in nature.
There will be seven characters so the answer is e
Answer:
V₁ = 96.2 mL
Explanation:
Given data:
Initial volume of NH₄OH required = ?
Initial molarity = 15.6 M
Final molarity = 3.00 M
Final volume = 500.0 mL
Solution:
Formula:
M₁V₁ = M₂V₂
M₁ = Initial molarity
V₁ = Initial volume of NH₄OH
M₂ =Final molarity
V₂ = Final volume
Now we will put the values.
15.6 M ×V₁ = 3.00 M×500.0 mL
15.6 M ×V₁ = 1500 M.mL
V₁ = 1500 M.mL /15.6 M
V₁ = 96.2 mL
