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Ivahew [28]
3 years ago
11

You have 400 kg of a radioactive substance with a short half-life of 1000 years. how much will be left after these times?

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
8 0
Half-life of a radioactive substance is the time required to reduce the amount of substance to half of its initial amount. 

In present case, half-life is material is given as 1000 years and initial amount of material is given as 400 kg
 
Answer 1) Since, half-life of radio-active substance is 1000 years, therefore after 1st half life, amount of the material will be left to half the initial amount. Hence, amount of substance left after 1000 years = 400/2 = 200 kg.

Answer 2) For 2000 years, radioactive material has crossed 2 times the half life. Therefore ,  amount of the material will be left to 1/4 the initial amount. Hence, amount of substance left after 2000 years = 400/4 = 100 kg.

Answer 3) 
 For 4000 years, radioactive material has crossed 4 times the half life. Therefore , amount of the material will be left to 1/16 the initial amount. Hence, amount of substance left after 4000 years = 400/16 = 25 kg.
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The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon
tigry1 [53]

Answer:

\boxed{\text{300 g}}

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:         32                          60

           CH₃OH + CO ⟶ CH₃COOH

m/g:        160

(a) Moles of CH₃OH

\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}

(b) Moles of CH₃COOH

\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}

(c) Mass of CH₃COOH

\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}

3 0
3 years ago
Which of the following phases of matter has a fixed shape and volume
kodGreya [7K]

A matter in the solid phase has a fixed shape and volume.

4 0
2 years ago
Read 2 more answers
Calculate 4 Srxn for this equation. Round to the
masya89 [10]

Answer:

On edge nuity, it is 79.

Explanation:

Please read the explanation.

Since u have no coefficients, this problem is super easy.

You just add your products, 115.5+69.91 = 185.41

Then add your reactants: 49.8+56.5= 106.3

Then subtract your reactants from your product and round to the nearest whole number.

106.3-185.41=79

It's really easy. Once u understand it, you can do any problem like this. If there is coefficients, just multiply the coefficient by the amount given to you in the problem.

7 0
2 years ago
What would be the bond order for He 2 2- molecule
LuckyWell [14K]

Answer:

In He2 molecule,

Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.

Molecular Orbitals thus formed are:€1s2€*1s2

It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .

Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2

Bond Order =Nb-Na/2

Bond Order =2-2/2=0

Since the bond order is zero so that He2 molecule does not exist.

Explanation:

8 0
3 years ago
The molar solubility of silver bromide, AgBr in pure water is 0.0007350 mol/L. What is the
gayaneshka [121]

Answer:

0.000000540

Explanation:

Step 1: Make an ICE chart for the solution of AgBr

"S" represents the molar solubility of AgBr

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I                           0             0

C                          +S          +S

E                           S             S

Step 2: Write the expression for the solubility product constant (Ksp)

Ksp = [Ag⁺] [Br⁻] = S × S

Ksp = S² = (0.0007350)² = 0.000000540

7 0
2 years ago
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