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xxTIMURxx [149]
3 years ago
7

4. Thorium-234 has a half-life of 24.1 days. What percentage of the original thorium-234 in a sample will remain after 144.6 day

s?
Chemistry
1 answer:
kari74 [83]3 years ago
3 0
How many half lives have elapsed?
actually the formula is like this
A=P(1/2)^(t/h)
P=iniital amount
t=time
h=half life
A=final amount
so

hmm, percent
so P=1 then
A=1(1/2)^(144.6/24.1)
A=1(1/2)^6
A=1/64
A=0.05625
so 5.625% is left
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John Archibald Wheeler

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Tron 200g dung dich NaCl 40% voi m gam dung dich NaCl 20% thu duoc dung dich NaCl 25% . tinh m
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8 0
2 years ago
density of a piece of ebony wood is 1.174 g/ml. what is the volume in quarts of a 2.1549 lb piece of this ebony wood?
lapo4ka [179]

0.88qt

Explanation:

Given parameters:

Density of ebony wood = 1.174g/ml

Mass of wood = 2.1549lb

Unknown:

Volume of the wood in quarts = ?

Solution:

Density is defined as the mass per unit volume of a substance.

Mass is the quantity of matter the wood contains.

Volume is the space it occupies.

To solve this problem, convert the mass form pounds into grams and plug into the density equation to find volume.

Then convert the volume from mL to quarts.

 Density = \frac{mass}{volume}

 Volume = \frac{mass}{density}

    1 pound = 453.592g

   2.1549 lb =  2.1549 x 453.592 = 977.5g

Volume = \frac{977.5}{1.174} = 832.6mL

     1 quart = 946.353mL

        Therefore 832.6mL x \frac{1qt}{946.353} = 0.88qt

learn more:

Density brainly.com/question/8441651

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7 0
3 years ago
Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is
weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane =  1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = <u>5.30 * 10³ kcal</u>

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = <u>3.28 * 10³ kcal</u>

<u />

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0
3 years ago
A red car and a blue car collide. What would be an example of Locard's exchange principle in this situation?
Lelu [443]

Answer:

The blue car left paint on the red car

Explanation:

8 0
3 years ago
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