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sesenic [268]
3 years ago
7

(6 points) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m

(t) of the remaining isotope Alice owns after t years.
Chemistry
1 answer:
Cloud [144]3 years ago
6 0

<u>Answer:</u> The equation to calculate the mass of remaining isotope is [A]=\frac{20}{10^{-0.217t}}

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = \ln 4=1.386yrs

Putting values in above equation, we get:

k=\frac{0.693}{1.386yrs}=0.5yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.5yr^{-1}

t = time taken for decay process

[A_o] = initial amount of the sample = 20 grams

[A] = amount left after decay process =  ? grams

Putting values in above equation, we get:

0.5=\frac{2.303}{t}\log\frac{20}{[A]}

[A]=\frac{20}{10^{-0.217t}}

Hence, the equation to calculate the mass of remaining isotope is [A]=\frac{20}{10^{-0.217t}}

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Strontium sulfate becomes less soluble in an aqueous solution when sodium sulfator is added because
horsena [70]

Answer:

The addition of sulfate ions shifts equilibrium to the left.

Explanation:

Hello!

In this case, according to the following ionization of strontium sulfate:

SrSO_3(s)\rightleftharpoons Sr^{2+}+SO_4^{2-}

It is evidenced that when sodium sulfate is added, sulfate, SO_4^{2+} is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:

The addition of sulfate ions shifts equilibrium to the left.

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7 0
3 years ago
Read 2 more answers
Which would exert more pressure on the floor, a 100kg man wearing snow shoes or a 60kg woman wearing high heels? PLEASE EXPLAIN
iris [78.8K]

Pressure can be defined as the force acting on a perpendicular surface per unit area.

Force exerted by a man of mass 100 kg wearing snow shoes = m.a

Where m = mass of the man = 100 kg

a = acceleration due to gravity= 9.8 m/s^{2}

Force exerted by the man of mass 100 kg = 100 kg(9.8 m/s^{2}) = 980 N

Force exerted by woman of mass 60 kg = 60 kg(9.8 m/s^{2}) = 588 N

Force exerted by 100 kg man is greater than that exerted as 60 kg woman. The area on which this force is acting determines the pressure. Pressure is inversely proportional to the area on which the force acts. Therefore, the pressure exerted by 100 kg man wearing snow shoes is less than the pressure exerted by a 60 kg woman woman wearing high heels as the force acts over a larger area when the man wears snow shoes when compared to the force exerted over a smaller area in case of the woman wearing high heels.

5 0
3 years ago
Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
4 years ago
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8 0
3 years ago
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liberstina [14]
A. hex- Normally with 6, the prefix is hex-. For example a polygon having 6 sides is known as hexagon.
8 0
4 years ago
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