Answer:
So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of
5.30
years, and are interested in finding how many grams of the sample would remain after
1.00
year and
10.0
years, respectively.
A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.
If you start with an initial sample
A
0
, then you can say that you will be left with
A
0
2
→
after one half-life passes;
A
0
2
⋅
1
2
=
A
0
4
→
after two half-lives pass;
A
0
4
⋅
1
2
=
A
0
8
→
after three half-lives pass;
A
0
8
⋅
1
2
=
A
0
16
→
after four half-lives pass;
⋮
Explanation:
now i know the answer
Answer:
Starch is a viable indicator in the titration process because it turns deep dark blue when iodine is present in a solution. When starch is heated in water, decomposition occurs and beta-amylose is produced
Answer:
2300J
Explanation:
1 kilojoule is 1000joules so to get how much is 2.3 multiply it with 1000
The reaction is
CaC₂(s) + 2H₂O (l) -----> Ca(OH)₂ (s) + C₂H₂ (g)
As we have data of gas ethyne (or acetylene), C₂H₂
We can calculate the moles of acetylene and from this we can estimate the mass of calcium carbide taken
the moles of acetylene will be calculated using ideal gas equation
PV =nRT
R = gas constant = 0.0821 Latm/molK
T = 385 K
V = volume = 550 L
P = Pressure = 1.25 atm
n = moles = ?
n = PV /RT = 1.25 X 550 / 0.0821 X 385 = 21.75 mol
As per balanced equation these moles of acetylene will be obtained from same moles of calcium carbide
moles of calcium carbide = 21.75mol
molar mass of CaC₂ = 40 + 24 = 64
mass of CaC₂ = moles X molar mass = 21.75 X 64 = 1392g