Explanation:
Half life is simply the amount of time it takes for half of a substance to decompose.
Options;
- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.
- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000
- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)
- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)
- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)
Answer:
6
Explanation:
You will see H6 and the H stands for helium and the 6 is how many of that atom is there
Answer:
Explanation:
Hello!
In this case, since the applied current for the 50.0 mins provides the following charge to the system:
As 1 mole of electrons carries a charge of 1 faraday, or 96,485 coulombs, we can compute the moles of electrons involved during the reduction:
Then the reduction of Ga³⁺ to Ga involves the transference of three electrons, we are able to compute the moles and therefore the mass of deposited gallium:
Best regards!
Answer:
Molarity = 1.93 mol.L⁻¹
Explanation:
Molarity is the unit of concentration used to specify the amount of solute in given amount of solution. It is expressed as,
Molarity = Moles / Volume of Solution ----- (1)
Data Given;
Mass = 11.3 g
Volume = 100 mL = 0.10 L
First calculate Moles for given mass as,
Moles = Mass / M.mass
Moles = 11.3 g / 58.44 g.mol⁻¹
Moles = 0.1933 mol
Now, putting value of Moles and Volume in eq. 1,
Molarity = 0.1933 mol ÷ 0.10 L
Molarity = 1.93 mol.L⁻¹
Answer:
- Brush off any chemical dust around the balance after each use
- Alert your TA or instructor about any issues with the balance
Explanation:
The best practices while sharing a lab balance with other students includes alerting your TA or instructor about issues with the balance and Brushing off any chemical dust around the balance after each use. the practices will help to avert/reduce the work hazards associated with working with a balance.
Brushing off any chemical dust after each use will reduce the chemical hazards in the lab while Alerting your instructor about any issues with the balance helps to eliminate mechanical hazards that a faulty balance can pose
