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klasskru [66]
3 years ago
5

Manganese-58 has a half-life of about 3 seconds. If you have a 90.0 gram sample, how

Chemistry
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

18.018 seconds.

Explanation:

Given that the half life of Manganese, Mn = 3 seconds. The initial sample mass = 90.0 gram, the final sample mass = 1.40 gram.

The general idea to the question is to look for the time it will take to decay from the initial mass that is 90 gram to 1.40 gram.

Therefore, we will be making use of the formula below;

J(t) = J(o) × (1/2)^t/t(hL).

Where t(hL) is the half life, t is the time taken, J(t)= mass after time,t and J(o) is the initial mass. So, let us slot in the values into the equation above.

1.4 = 90 × (1/2)^ t/3.

1.4/90 = (1/2)^t/3.

t/3 = log(0.5) (1.4/90).

+Please note that the 0.5 of the log is at the subscript).

That is the base 0.5 logarithm of (1.4/90) 0.01556 is 6.0060141295.

t = 3 × 6.0060141295.

t = 18.018 seconds.

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Bdjdisisididiieddiidid
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2 years ago
if a sample of gas at 25.2 c has a volume of 536mL at 637 torr, what will its volume be if the pressure is increased to 712 torr
nignag [31]
Considering ideal gas:
PV= RTn

T= 25.2°C = 298.2 K

P1= 637 torr = 0.8382 atm

V1= 536 mL = 0.536 L

:. R=0.082 atm.L/K.mol

:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))

:. n= O.0184 mol

Then,
P2= 712 torr = 0.936842 atm

V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)

:.V2 = 0.4796 L
OR
V2 = 479.6 ml

6 0
3 years ago
This transition metal is in group 12 and has less than 34 protons
slava [35]

Zinc (Zn) has less than 34 protons, 30 to be exact, and is a transition metal in Group 12. Note: it is also called a "post-transition metal."

4 0
2 years ago
If a substance has a half life of 58 years and starts with 500 g radioactive, how much remains radioactive after 30 years?
Vilka [71]

Answer:

A = 349 g.

Explanation:

Hello there!

In this case, since the radioactive decay kinetic model is based on the first-order kinetics whose integrated rate law is:

A=Ao*exp(-kt)

We can firstly calculate the rate constant given the half-life as shown below:

k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{58year}=0.012year^{-1}

Therefore, we can next plug in the rate constant, elapsed time and initial mass of the radioactive to obtain:

A=500g*exp(-0.012year^{-1} *30year)\\\\A=349g

Regards!

5 0
3 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

3 0
2 years ago
Read 2 more answers
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