The mass can be determined with the formula m=800(.5)^(t/5)
-66.7
The relevant equation behind this problem is Q=m*c* ΔT Where Q is the energy that must be added to or taken from the system, m is the mass of the object, c is the objects specific heat, and ΔT is the change in temperature in Celsius or Kelvin. Plugging in the given values we get that Q=.015kg * 128J/(kg*C) * 10C=19.2J. Therefore, you need 19.2 joules of heat in order to raise the temperature of a .015kg sample of lead by 10 degrees Celsius.
Q (heat) = mass*Specific Heat*Temperature differenceI assume 2.5 * 101 Joules? or 25 J & 10 grams of substance?25 J = 10 grams * (SH) * (70-10oC)25 J =10 g *60oC * SHSH = 25 J/(10 g * 60oC)S.H. = 0.0417 J/gOC
Pi is irrational. This means that as the extent of this number is infinite, therefore any numbers present in this number for an unspecified amount of time are also considered infinite. However, this is not necessarily the case. Consider that pi could go on forever, but after the billionth digit, for example, there would be no 3's.This could easily be the case, but we would never know for certain, because it doesn't matter how many digits we discover not to be 3's, the next digit could always be any number, including a 3.However, there are exactly digits, and provided these digits are positioned randomly throughout the number, (which is what irrational means) there is a 1 in 10 chance of the next digit being a 3. This also means that in any given sample of consecutive digits of pi, 1/10 of the digits are likely to be a 3. The bigger the sample, the closer to 10% that the number of 3's will be. So if we take a million digits of pi, we can estimate that approximately 100,000 of them would be 3's. We can check this estimate by looking at how many threes there really are: 100,230 3's according to various sources.
Depends on how much of the sample there is.
Cobalt-60 has two gamma photopeaks at 1.17 MeV and 1.33 MeV. Radon-222 is an alpha emitter. Bismuth-83 is not a valid isotope. 83 is the atomic number of bismuth, but you need to know the atomic mass number. The nuclide with the longest half-life is bismuth-208, and it decays by beta+ decay.
You forgot to say that isotope is.
2.09
5.0 mg is the total mass of 222Rn remaining in an original 160-milligram sample of 222Rn after 19.1 days.
yes
Density of a substance = (mass of a sample of the substance) divided by (volume of the same sample)
<p>Sample sugar<p>
=AVERAGE(A1:A34)
The following is an experiment which can be carried out in a laboratory.ReagentsSolid bismuth chlorideConcentrated hydrochloric acidMethodologyTake approximately 1 gram of bismuth chloride to a test tube. And then add around 2 mL of water into it. Now, add the concentrated hydrochloric acid drop-wise till the solid dissolves.Next add excess water till a white suspension is formed.Then get a sample from the suspension and add the concentrated hydrochloric acid drop-wise.Follow the steps 2 and 3 alternately.The white suspension is formed by bismuth oxychloride.Addition of dilute acid will not give the proper results as it results in a lack of the concentration of chloride ions.
sample area/standard area*standard weight/sample weight*standard purity/100*100
The chemical symbol (not formula, macroscopic sample) of neon is Ne.