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2 years ago

Need answer now! 100 points

Chemistry

2 answers:

ratelena [41]2 years ago

5 0

Answer: D

divide the moles of the substances given in the question by the co-efficient in the equation. Smallest quotient is the limiting reactant

s344n2d4d5 [400]2 years ago

3 0

Answer:

D) HBr

Explanation:

Given data:

Number of moles of Al = 8 mol

Number of moles of HBr = 8 mol

Limiting reactant = ?

Solution:

2Al + 6HBr → 2AlBr₃ + 3H₂

now we will compare the moles of reactants with products.

Al : AlBr₃

2 : 2

8 : 8

Al : H₂

2 : 3

8 : 3/2×8 = 12

HBr : AlBr₃

6 : 2

8 : 2/6×8 = 2.67

HBr : H₂

6 : 3

8 : 3/6×8 = 4

HBr produced less number of moles of product thus it will act as limiting reactant.

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The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

$\frac{dA}{dt}= kA$

$\Rightarrow \frac{dA}{A}=kdt$

Integrating both sides

ln A= kt+c₁

$\Rightarrow A= e^{kt+c_1}$

$\Rightarrow A=Ce^{kt}$ [ $e^{c_1}=C$ ]

The initial condition is A(0)= A₀,

$\therefore A_0=Ce^{0.k}$

$\Rightarrow C=A_0$

$\therefore A=A_0e^{kt}$ .........(1)

Given that the $Pb_{209}$ has half life of 3.3 hours.

For half life $A=\frac12 A_0$ putting this in equation (1)

$\frac12A_0=A_0e^{k\times3.3}$

$\Rightarrow ln(\frac12)= 3.3k$ [taking ln both sides, $ln \ e^a=a$ ]

$\Rightarrow k=\frac{ln \frac12}{3.3}$

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

$\therefore 0.2=1e^{(-0.21)\times t}$

$\Rightarrow e^{-0.21t}=0.2$

$\Rightarrow -0.21t= ln(0. 2)$

$\Rightarrow t= \frac{ln (0.2)}{-0.21}$

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

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