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Lina20 [59]
3 years ago
12

At the beginning of an experiment, a scientist has 176 grams of radioactive goo. After 165 minutes, her sample has decayed to 5.

5 grams.
What is the half-life of the goo in minutes?

Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?

How many grams of goo will remain after 50 minutes? ...?
Chemistry
1 answer:
Paul [167]3 years ago
6 0
The half-life equation is written as:

An = Aoe^-kt

We use this equation for the solution. We do as follows:

5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes? 
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE


Find a formula for G(t) , the amount of goo remaining at time t.G(t)=? 

G(t) = 176e^-0.02t

How many grams of goo will remain after 50 minutes? 

G(t) = 176e^-0.02(50) = 64.75 g
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7 0
4 years ago
if you mixed 72.9g hydrochloric acid(aq) with 150g silver acetate(aq), what would be the limiting reagent?​
horsena [70]

Answer:

                      Silver Acetate would be the Limiting Reagent.

Explanation:

                    The balance chemical equation for the given double displacement reaction is as;

                            HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂

Step 1: <u>Calculate Moles of Starting Materials:</u>

Moles of HCl:

                      Moles  =  Mass / M.Mass

                      Moles  =  72.9 g / 36.46

                      Moles =  1.99 moles

Moles of AgC₂H₃O₂:

                      Moles  =  150 g / 166.91 g/mol

                      Moles  =  0.898 moles

Step 2: <u>Find out Limiting reagent as:</u>

According to balance chemical equation.

              1 mole of HCl reacts with  =  1 mole of AgC₂H₃O₂

So,

         1.99 moles of HCl will react with  =  X moles of AgC₂H₃O₂

Solving for X,

                     X =  1.99 mol × 1 mol / 1 mol

                     X =  1.99 mol of AgC₂H₃O₂

Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.

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