Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K
Answer:
See explanation and image attached
Explanation:
The reaction is an E2 reaction. It is a synchronous reaction.
The base KOC(CH3)3 abstracts a proton as the bromide ion leaves in a single step.
This yields the product as shown in the image attached.
<span>The products of the combustion of a hydrocarbon are dependent on if the reaction is run to completion, therefore, you require sufficient Oxygen to completely combust the hydrocarbon. If run to completion the products are H20 and CO2. If not then some CxHy and some O2 are left.</span>
The number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.
The given parameters;
- <em>mass of the reacting sodium = 0.231 g</em>
- <em>1 mole of an atom = 6.02 x 10²³ atoms</em>
- <em>the atomic mass of sodium is 23 g/mol</em>
The number of moles of 0.231 g of sodium available is calculated as;
The number of atoms of 0.01 mole of sodium available is calculated as;
1 mole ---------- 6.02 x 10²³ atoms
0.01 ----------------- ?
= 0.01 x 6.02 x 10²³ atoms
= 6.02 x 10²¹ atoms.
Thus, the number of atoms present in 0.231 g of sodium is 6.02 x 10²¹ atoms.
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