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Paladinen [302]
3 years ago
14

Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person or animal that ingests it.

The median lethal dose of KCN for a person weighing 145 lb (65.8 kg ) is 7.27 x 10^−3 mol .
1. What volume of a 0.0820 M KCN solution contains 7.27x 10^−3 mol of KCN?
Chemistry
1 answer:
Artemon [7]3 years ago
4 0

Answer:

88,7 mL of solution

Explanation:

Molarity (Represented as M) is an unit of chemical concentration that is defined as the ratio between moles of solute per liters of solution, that is:

Molarity = moles of solute / Liters of solution

If molarity of KCN solution is 0,0820M and moles of KCN are 7,27x10⁻³ moles:

0,0820M = 7,27x10⁻³ moles / Liters of solution

Liters of solution = 0,0887L = <em>88,7 mL of solution</em>

I hope it helps!

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Answer:

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Explanation:

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How many atoms of gold are in gold nugget with a mass of 4.25 g
Feliz [49]
Number of Atoms in Gold for given mass can be calculated using following formula,

              # of Moles  =  Number of Atoms / 6.022 × 10²³
Or,
             Number of Atoms  =  Moles × 6.022 × 10²³     ------- (1)    

Calculating Moles,
As,
                              Moles  =  Mass / M.mass
So,
                              Moles  =  4.25 g / 196.96 g/mol

                              Moles  =  0.0215

Putting value of mole in eq.1,

             Number of Atoms  =  0.0215 × 6.022 × 10²³

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Result:
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Calculate the ph of a 0.17 m solution of c6h5nh3no3 (kb for c6h5nh2 = 3.8 x 10-10). record your ph value to 2 decimal places.
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3 years ago
A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin
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Answer:

12.371 g

Explanation:

Given :

m_{evaporating\ dish}=1.135\ g

m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g

m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g

m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g

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m_{evaporating\ dish}=1.135\ g

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m_{evaporating\ dish}=1.135\ g

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m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g

Mass of water:

m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g

m_{water}=12.371\ g

4 0
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