Amount of Niobium-91 initially
= 300/91 =3.2967mol
2040 years = 3 ×680 = 3 half-lives
therefore, amount left = 0.4121mol
mass of Niobium-91 remaining = 0.4121 ×91 =37.5g
Answer:
The correct answer is option d.
Explanation:
On submission of an incorrect answer to a multiple-choice question with n options, we will lose:
of credit
If we multiple-choice question has five answer choices. And we have submitted 1 wrong answer. the fraction of credit we loose :
of credit
of credit
credit
The percentage of credit we loose:
of credits
of credits
210 cm converted into inches is 82.67716535 in.
Answer:
If you constructed graphs, what trends do they indicate in your data?
Correct Question :
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80°C
In an experiment to determine the specific heat of a metal student transferred a sample of the metal that was heated in boiling water into room temperature water in an insulated cup. The student recorded the temperature of the water after thermal equilibrium was reached. The data we shown in the table above. Based on the data, what is the calculated heat absorbed by the water reported with the appropriate number of significant figures?
Answer:
1642 J
Explanation:
Given:
Mass of water = 50.003g
Temperature of water= 24.95C
Specific heat capacity for water = 4.184J/g C
Mass of metal = 63.546 g
Temperature of metal 99.95°C
Specific heat capacity for metal ?
Final temperature = 32.80° C
To calculate the heat absorbed by water, Q, let's use the formula :
Q = ∆T * mass of water * specific heat
Where ∆T = 32.80°C - 24.95°C = 7.85°C
Therefore,
Q= 7.85 * 50.003 * 4.184
Q = 1642.32 J
≈ 1642 J
