<u>Answer: </u>The correct answer is Silver.
<u>Explanation:</u>
Specific heat of fusion is defined as the amount of heat which is required to raise the temperature of 1 gram of a substance to 1°C. It is generally expressed in kJ/mol
We are required to find the substance which require more heat. For that we need to know the specific heat of all the substances.
The substance which have the highest specific heat, will require more heat.
The specific heat of the given substances are:
Silver = 11.3 kJ/mol
Sulfur = 1.7175 kJ/mol
Water = 5.98 kJ/mol
Lead = 4.799 kJ/mol
The specific heat of silver is the highest and hence, will require more heat.
Hence, the correct answer is silver.
Answer:
C. BF3
Explanation:
The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.
Answer:
The volume is 1.2L
Explanation:
Initial volume (V1) = 700mL = 0.7L
Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K
Initial pressure = 106.6kPa = 106600Pa
Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K
Final pressure (P2) = 66.6kPa = 66600Pa
Final volume (V2) = ?
To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.
(P1 × V1) / T1 = (P2 × V2) / T2
solve for V2 by making it the subject of formula,
P1 × V1 × T2 = P2 × V2 × T1
V2 = (P1 × V1 × T2) / (P2 × T1)
V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)
V2 = 22397193 / 18657990
V2 = 1.2L
The final volume of the gas is 1.2L
A contains 38.5 g of tin for each 12.3 g of fluorine:
<span>mole ratio: </span>
<span>(38.5 g)/(118.71 g/mol):(12.3 g)/(18.998 g/mol) = 0.324:0.647 = 1:2 ⇒ SnF₂ </span>
<span>B contains 56.5 g of tin for each 36.2 g of fluorine: </span>
<span>mole ratio: </span>
<span>(56.5 g)/(118.71 g/mol):(36.2 g)/(18.998 g/mol) = 0.476:1.905 = 1:4 ⇒ SnF₄
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Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
