The heat of the reaction, in kJ, when 4.18 g of the hydrocarbon are combusted 775.70 kJ.
The heat energy is given as :
q = m c ΔT + Ccal ΔT
q = ( 974 g× 4.184 ×6.9) + 624 ×6.9
q = 32424.59 J
moles of hydrocarbon = 0.0418 mol
heat of combustion = 32424.59 J / 0.0418 mol
= 775707.89 J
= 775.70 kJ
Thus, A 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water increases by 6.9 °C when the hydrocarbon is combusted. the calorimeter constant for the calorimeter was determined to be 624 J/°C. what is the heat of the reaction is 775.70 kJ.
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Answer:
arms
Explanation:
because arms can be metric system of our body
Answer is (2) - hydrogen carbonate
<em>Explanation:</em>
NaHCO₃ is an ionic compound which is made from Na⁺ and HCO₃⁻ ions. The decomposition is
NaHCO₃ → Na⁺ + HCO₃⁻
Among the resulted ions, Na⁺ is a monatomic ion while HCO₃⁻ is a polyatomic ion.
<em>Polyatomic ions mean ions which are made of two or more different atoms.</em>
HCO₃⁻ is made from 3 atoms as H, C and O. The name of HCO₃⁻ ion is bicarbonate or hydrogen carbonate.
<span>Assuming ideal gas, we can label this system to follow the Boyle's Law. This Law can be applied to systems held at constant temperature. The formula used is: PV=k, where k is a constant. From the formula, P and V are inversely proprotional. So, if you graph P vs V, the graph would start from the top, then curves down towards the right. </span>
