Answer- The particles in a solid gain enough energy to overcome the bonding forces holding them firmly in place. Typically, during melting, the particles start to move about, staying close to their neighbouring particles, then move more freely.
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:
0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Explanation:
Mass percentage of oxalic acid = 5.50%
This means that in 100 grams of solution there are 5.50 grams of oxalic acid.
Mass of solution , m = 100
Volume of the solution = V
Density of the solution = d = 1.024 g/mL

V = 97.66 mL = 0.09766 L
(1 mL = 0.001 L)
Moles of oxalic acid = 

The molarity of the solution :

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.
Answer:
you need to use the 2 because I already did it
Explanation:
db
substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>