A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead
1 answer:
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The reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
<h3>What is metal displacement reaction? </h3>Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.
As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.
Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.
Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
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Answer:
The standard enthalpy of formation of HgO is -90.7 kJ/mol.
Explanation:
The reaction between Hg and oxygen is as follows.
From the given,
Molar mass of HgO = 216.59 g/mol
Mass of HgO decomposed = 18.5 g
Amount of heat absorbed = 7.75 kJ
From the reaction,
The standard enthalpy of formation =
During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.
For the formation of 1 mol of HgO , 90.7 kJ of energy is release
Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol
<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For NaBr:</u>
Given mass of NaBr = 11.1 g
Molar mass of NaBr = 103 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of 1-butanol and NaBr is:
By Stoichiometry of the reaction
1 mole of NaBr produces 1 mole of 1-bromobutane
So, 0.108 moles of NaBr will produce = moles of 1-bromobutane
- Now, calculating the mass of 1-bromobutane from equation 1, we get:
Molar mass of 1-bromobutane = 137 g/mol
Moles of 1-bromobutane = 0.108 moles
Putting values in equation 1, we get:
- To calculate the percentage yield of 1-bromobutane, we use the equation:
Experimental yield of 1-bromobutane = 7.2 g
Theoretical yield of 1-bromobutane = 14.80 g
Putting values in above equation, we get:
Hence, the percent yield of the 1-bromobutane is 48.65 %