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.
Answer:
Explanation:
Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²
2x 0.007 50 + x
![K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BAg%24%5E%7B%2B%7D%24%5D%24%5E%7B2%7D%24%5BCO%24_%7B3%7D%5E%7B2-%7D%24%5D%7D%20%3D%20%282x%29%5E%7B2%7D%5Ctimes%200.00750%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5C0.0300x%5E%7B2%7D%20%3D%208.10%20%5Ctimes%2010%5E%7B-12%7D%5C%5Cx%5E%7B2%7D%20%3D%202.70%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.70%20%5Ctimes%2010%5E%7B-10%7D%7D%20%3D%20%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20maximum%20concentration%20of%20Ag%24%5E%7B%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.64%5Ctimes%2010%5E%7B-5%7D%7D%5Ctextbf%7B%20mol%2FL%20%7D%7D%24%7D)
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
Answer:
Ley.
Explanation:
En la teoría de la ciencia, la regularidad de los procesos en la naturaleza se denomina ley de la naturaleza. Las leyes naturales se diferencian de otras leyes en que los seres humanos no pueden ponerlas en vigor ni anularlas a su discreción. En tal sentido, la composición química del agua es indudablemente una ley natural, en tanto el hombre no puede modificarla sin modificar las características inherentes del agua como tal.
Answer:
Buffers are solutions that resist changes in pH, upon addition of small amounts of acid or base. The can do this because they contain an acidic component, HA, to neutralize OH- ions, and a basic component, A-, to neutralize H+ ions. Since Ka is a constant, the [H+] will depend directly on the ratio of [HA]/[A-].
hope it's help
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