<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>
Explanation:
- Water has high specific heat due to hydrogen bonds present in it.
- The Ionisation of water does not affect the specific heat of the water.
- On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
- On increasing the temperature, there is the breaking of bonds hence option (b) is correct.
The component of the candle burning in the surrounding has been the oxygen in the air.
The burning of candle wax and wick has been the chemical reaction. It has been based on the reaction of wick with the atmospheric oxygen, resulting in the formulation of the wax burning.
<h3>Chemical reaction of burning of wax</h3>
The wax has been vaporizes by the heat of the flame, that has been resulted by the burning. The wick has been able to react with the oxygen and form the byproducts that helps in flame burning.
The end products have been wick and oxygen as the wax has been consumed in the reaction. The air in the surrounding has oxygen as the part of the system, as it has been involved in the reaction.
Learn more about candle burning, here:
brainly.com/question/25955977
To prevent the hydrolysis and to catalyse the reaction.
Explanation:
- Sulphuric acid is the catalyst and also a dehydrating agent in this reaction.
- Sulphuric acid is using in redox reaction because sulphuric acid is providing H+ ions which is necessary for this reaction to occur more quickly, but the sulphate ions from the sulphuric acid barely react during this process. So H2SO4 is adding in this reaction to make it more acidic.
- H2SO4 is preventing hydrolysis by providing excess H+ ions into the reaction. H2SO4 is stable towards the direction of oxidation.
Answer:
187.34 atm
Explanation:
From the question,
PV = nRT.................. Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
make P the subject of the equation
P = nRT/V.............. Equation 2
n = mass(m)/molar mass(m')
n = m/m'............... Equation 3
Substitute equation 3 into equation 2
P = (m/m')RT/V............ Equation 4
Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L
Constant: m' = 2 g/mol, R = 0.082 atmL/K.mol
Substitute these values into equation 4
P = (46/2)(0.082×298)/3
P = (23×0.082×298)/3
P = 187.34 atm
here it is! mL to Liters to moles to molecules