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Reptile [31]
4 years ago
13

Approximately what precent of parent isotopes remains after 0.5 half lives have passed

Chemistry
1 answer:
Charra [1.4K]4 years ago
5 0
If 1 half life is over than 50 percent is left
I think it would be logicak to assume after half a half life passes there will be 75 percent left
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A chemist studies the reaction below. 2NO(g) + Cl2(g) 2NOCl(g) He performs three experiments using different concentrations and
Dmitry_Shevchenko [17]

Answer:

1. Rate =k [NO]^{2}[Cl_{2}]

2. k= 0.42 \frac{L^{2}}{mol^{2}*s}

Explanation:

Rate =k [NO]^{m}[Cl_{2}]^{n}

Rate1 = k[0.4]^{m}[0.3]^{n}=0.02\\Rate 2=k [0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate1}{Rate2}=\frac{0.02}{0.08} =\frac{k[0.4]^{m}[0.3]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{1}{4} =(\frac{1}{2} )^{m},\\m=2

Rate3 =k [0.8]^{m}[0.6]^{n}=0.16\\Rate 2= k[0.8]^{m}[0.3}]^{n}=0.08\\\\\frac{Rate3}{Rate2}=\frac{0.16}{0.08} =\frac{k[0.8]^{m}[0.6]^{n}}{k[0.8]^{m}[0.3]^n}} \\\\\frac{2}{1} =(\frac{2}{1} )^{n},\\n=1

Rate =k [NO]^{2}[Cl_{2}]^{1}

Rate =k [NO]^{2}[Cl_{2}]^{1}\\Rate 1=k [0.4]^{2}[0.3]^{1} =0.02\\k*0.16*0.3=0.02\\k=\frac{0.02}{0.16*0.3}=\frac{1}{8*(\frac{3}{10} )}=\frac{5}{12}  = 0.42 \frac{L^{2}}{mol^{2}*s}

6 0
3 years ago
What is the percent yield of 20.0g P4 O10 are produced?
enyata [817]

Answer:

You must divide the grams of your actual yield by the grams of the theoretical yield and multiply by 100 in order to obtain percent yield

Explanation:

7 0
3 years ago
6b<br> Which of the following is a starting compound during<br> cellular respiration?
Rus_ich [418]

Answer: Oxygen and glucose.

Explanation:

Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.Jun 1, 2020

3 0
3 years ago
Brainliest for an answer!
Degger [83]

The volume of H₂ : = 15.2208 L

<h3>Further explanation</h3>

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

5 0
3 years ago
What would you observe in a solution of 30 g of KC1O3 in 100 g of water at 10 c?
Mrac [35]

Answer:

Option A is correct. About 5 g of the KClO3 is dissolved

Explanation:

KClO3 is not very good soluble in water.

So, Option C is impossible, because KClO3 is poorly soluble in water.

The low solubility of KClO3 in water causes KClO3 to isolate itself from the reaction mixture by precipitating out of solution.

So, option D will either happen.There will be a part of KClO3 dissolve.

At 10 °C, KClO3 has a solubility of 4.46 g/100 gram (10 °C).

Option A is correct. About 5 g of the KClO3 is dissolved

7 0
3 years ago
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