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Ronch [10]
3 years ago
7

uranium-232 has a half-life of 68.8 years. after 344.0 years how much uranium-232 will remain from a 125.0g sample

Chemistry
2 answers:
zhuklara [117]3 years ago
5 0

B. 3.13 g

Hope this helps

GaryK [48]3 years ago
4 0

Answer:

4 grams

Explanation:

A = A₀e^⁻kt

A₀ = 125.0 grams

k= 0.693/t(1/2) = (0.693/68.8) yrs⁻¹ = 0.01 yrs⁻¹

t = 344.0 years

A = 125.0g·[e^-(0.01yrs⁻¹)(344.0yrs)] = 125(0.032)grams = 4.000g (4 sog. figs. based on A₀ = 125.0 grams)

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A lake has been infected by some type of new algae that is unknown. Every single day the amount of surface area that the algae t
FrozenT [24]

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It takes 86 days take to cover half of the lake

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In the day #3 is 2*2*X = X*2²

...

In the day #n the amount of the algae is X*2^(n-1)

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The time transcurred is:

3.87x10²⁵m³ = 1m³*2^(n-1)

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4 0
3 years ago
Magnesium metal (0.100 mol) and a volume of aqueous hydrochloric acid that contains 0.500 mol of HCl are combined and react to c
jok3333 [9.3K]

Answer:

2.24 L of hydrogen gas, measured at STP, are produced.

Explanation:

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Moles of hydrochloric acid, HCl = 0.500 mol

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Mg_{(s)} + 2HCl_{(aq)}\rightarrow MgCl_2_{(aq)} + H_2_{(g)}

1 mole of Mg reacts with 2 moles of HCl

0.100 mol of Mg reacts with 2*0.100 mol of HCl

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Limiting reagent is the one which is present in small amount. Thus, Mg is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of Mg on reaction forms 1 mole of H_2

0.100 mole of Mg on reaction forms 0.100 mole of H_2

Mole of H_2 = 0.100 mol

At STP,  

Pressure = 1 atm  

Temperature = 273.15 K

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V L = 0.100 × 0.0821 L.atm/K.mol × 273.15 K  

<u>⇒V = 2.24 L</u>

2.24 L of hydrogen gas, measured at STP, are produced.

4 0
3 years ago
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