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3 years ago

If you start with 1.000 gram sample of the isotope how much time would pass before you have just 0.100 grams of the isotope left

Chemistry

1 answer:

Vadim26 [7]3 years ago

6 0

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

, then you can say that you will be left with

→

after one half-life passes;

⋅

→

after two half-lives pass;

⋅

→

after three half-lives pass;

⋅

→

after four half-lives pass;

⋮

Explanation:

now i know the answer

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9. Gasoline burns easily if ignited

Elena L [17]

Answer:

I think it is chemical change

5 0

3 years ago

A sample of gas at STP has a volume of 5.23 L. If the gas volume is changed to 3.45 L at 293 K, calculate the new pressure of th

Sever21 [200]

Answer: P=1.63atm

Explanation:

Stp means standard temperature and pressure

Standard temperature =273k

Standard pressure =1atm

Using the formula

P1V1/T1=P2V2/T2

P1= 1atm

V1=5.23L

T1=273k

P2=?

V2=3.45L

T2=293k

Substitute the values

1×5.23/273=p2×3.45/293

Cross multiply

293×1×5.23=p2×3.45×273

1532.39=941.85p2

P2=1532.39/941.85

P2=1.627

P2=1.63atm

3 0

3 years ago

Given the following information: benzoic acid = C6H5COOH hydrocyanic acid = HCN C6H5COOH is a stronger acid than HCN (1) Write t

Maurinko [17]

Answer:

The net ionic equation is

C6H5COOH+ CN-= C6H5COO- + HCN

Explanation:

From the ionic equation

C6H5COOH + Na+ + CN- = C6H5COO- + Na+ + HCN

Only sodium is the spectator ion, so it cancels out, since C6H5COOH and HCN do not ionize completely they are left undissociated

5 0

3 years ago

What is the boiling point of a solution produced by adding 610 g of cane sugar (molar mass 342.3 g/mol) to 1.4 kg of water? For

wel

Answer:

Boiling point of solution is $100.65^{0}\textrm{C}$

Explanation:

Cane sugar is a non-volatile solute.

According to Raoult's law for a non-volatile solute dissolved in a solution-

$\Delta T_{b}=K_{b}.m$

Where, $\Delta T_{b}$ is elivation in boiling point of solution, $K_{b}$ is ebbulioscopic constant of solvent (how much temperature is raised for dissolution of 1 mol of non-volatile solute) and m is molality of solution.

Here, $K_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}$

610 g of cane sugar = $\frac{610}{342.3}$ moles of cane sugar

= 1.78 moles of cane sugar

So, molality of solution (m) = $\frac{1.78}{1.4}mol.kg^{-1}=1.27mol.kg^{-1}$

Plug in all the values in the above equation, we get-

$\Delta T_{b}=0.51^{0}\textrm{C}.kg.mol^{-1}\times 1.27mol.kg^{-1}=0.65^{0}\textrm{C}$

So, boiling point of solution = $(100+0.65)^{0}\textrm{C}=100.65^{0}\textrm{C}$

7 0

3 years ago

What is the ph of a buffer consisting of 0.200 m hc2h3o2 and 0.200 m kc2h3o2? the k a for hc2h3o2 is 1.8×10−5. view available hi

AURORKA [14]

PH = pKa + log $\frac{[base]}{[Acid]}$
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
so pH = 4.74 + log (0.2/0.2) = 4.74
This is called maximum buffer capacity (when acid conc. and base conc. are equal) the pH = pKa in this case

6 0

3 years ago