Half-life of a radioactive substance is the time required to reduce the amount of substance to half of its initial amount.
In present case, half-life is material is given as 1000 years and initial amount of material is given as 400 kg
Answer 1) Since, half-life of radio-active substance is 1000 years, therefore after 1st half life, amount of the material will be left to half the initial amount. Hence, amount of substance left after 1000 years = 400/2 = 200 kg.
Answer 2) For 2000 years, radioactive material has crossed 2 times the half life. Therefore , amount of the material will be left to 1/4 the initial amount. Hence, amount of substance left after 2000 years = 400/4 = 100 kg.
Answer 3) For 4000 years, radioactive material has crossed 4 times the half life. Therefore , amount of the material will be left to 1/16 the initial amount. Hence, amount of substance left after 4000 years = 400/16 = 25 kg.
Answer: “We live on a hunk of rock and metal that circles a humdrum star that is one of 400 billion other stars that make up the Milky Way Galaxy which is one of billions of other galaxies which make up a universe which may be one of a very large number, perhaps an infinite number, of other universes. That is a perspective on human life and our culture that is well worth pondering.”
― Carl Saga
Is this it mate?
Answer:
44.2 L
Explanation:
Use Charles Law:
We have all the values except for V₂; this is what we're solving for. Input the values:
- make sure that your temperature is in Kelvin
From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:
Therefore, V₂ = 44.2 L
It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
