Answer:
Explanation:
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In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:
Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:
Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:
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Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
Vinegar pH 3.2: Weak acid
Battery acid pH 0.5: Strong acid
Shampoo pH 7.0: Neutral
Ammonia pH 11.1 Strong base
The initial temperature of the copper metal was 27.38 degrees.
Explanation:
Data given:
mass of the copper metal sample = 215 gram
mass of water = 26.6 grams
Initial temperature of water = 22.22 Degrees
Final temperature of water = 24.44 degrees
Specific heat capacity of water = 0.385 J/g°C
initial temperature of copper material , Ti=?
specific heat capacity of water = 4.186 joule/gram °C
from the principle of:
heat lost = heat gained
heat gained by water is given by:
q water = mcΔT
Putting the values in the equation:
qwater = 26.6 x 4.186 x (2.22)
qwater = 247.19 J
qcopper = 215 x 0.385 x (Ti-24.4)
= 82.77Ti - 2019.71
Now heat lost by metal = heat gained by water
82.77Ti - 2019.71 = 247.19
Ti = 27.38 degrees
