Heat is usually transferred by convection
Dihydrogen oxide is the right answer. Dihydrogen oxide is just 2 hydrogen and 1 oxygen which is H2O or water.
1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?
Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.
2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.
3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.
What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.
<span>6.50x10^3 calories.
Now we have 4 pieces of data and want a single result. The data is:
Mass: 100.0 g
Starting temperature: 25.0°C
Ending temperature: 31.5°C
Specific heat: 1.00 cal/(g*°C)
And we want a result with the unit "cal". Now you need to figure out what set of math operations will give you the desired result. Turns out this is quite simple. First, you need to remember that you can only add or subtract things that have the same units. You may multiply or divide data items with different units and the units can combine or cancel each other. So let's solve this:
Let's start with specific heat with the unit "cal/(g*°C)". The cal is what we want, but we'ld like to get rid of the "/(g*°C)" part. So let's multiply by the mass:
1.00 cal/(g*°C) * 100.0 g = 100.0 cal/°C
We now have a simpler unit of "cal/°C", so we're getting closer. Just need to cancel out the "/°C" part, which we can do with a multiplication. But we have 2 pieces of data using "°C". We can't multiply both of them, that would give us "cal*°C" which we don't want. But we need to use both pieces. And since we're interested in the temperature change, let's subtract them. So
31.5°C - 25.0°C = 6.5°C
So we have a 6.5°C change in temperature. Now let's multiply:
6.5°C * 100.0 cal/°C = 6500.0 cal
Since we only have 3 significant digits in our least precise piece of data, we need to round the result to 3 significant figures. 6500 only has 2 significant digits, and 6500. has 4. But we can use scientific notation to express the result as 6.50x10^3 which has the desired 3 digits of significance. So the result is 6.50x10^3 calories.
Just remember to pay attention to the units in the data you have. They will pretty much tell you exactly what to add, subtract, multiply, or divide.</span>
The mass of one atom of Carbon is 2.0×10^−26 kg.
The mass of the period mark(kg)= 1.0x10^-7 kg.
# of carbon atoms=1x10^-7/2x10^-26
# of carbon atoms= 5x10^-34
