The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.
The concentration of the Nitric acid solution : 0.114 M
<h3>Further explanation </h3>
Titration is a procedure for determining the concentration of a solution (analyte) by reacting with another solution whose known concentration (usually a standard solution) is called the titrant. Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range
Titrations can be acid-base titration, depositional titration, and redox titration. An acid-base titration is the principle of neutralization of acids and bases
Reaction
HNO₃ + NaOH → NaNO₃ + H₂O
Concentration a standard solution of sodium hydroxide : 0.0998 mol/dm³ , and the volume = 25 cm³
moles NaOH=
<em>From the equation, mol ratio HNO₃ : NaOH = 1 : 1, so mol HNO₃ = mol NaOH=</em><em>2.495 mlmoles</em>
<em></em>
The volume of HNO₃ = 21.8 cm³, so the concentration :
Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:
Now, find moles of 
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
<span>number of moles are the amount of substance of an element .
number of moles can be calculated as follows;
number of moles = mass present / molar mass of element
molar mass of Ar - 40 g/mol
number of moles of Ar = 22 g / 40 g/mol = 0.55 mol .
there are 0.55 mol of Ar</span>
