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cupoosta [38]
3 years ago
6

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 185 lb (83.9 kg ) is 8.50×10−3 mol . what volume of a 0.0320 m kcn solution contains 8.50×10−3 mol of kcn?
Chemistry
1 answer:
skelet666 [1.2K]3 years ago
7 0
<span>(8.5 x 10^-3 mol) / (0.0320 mol/L KCN)=0.265625 L=265.625 ml. This is the volume of a 0.0320 M KCN solution that would give a median lethal dose to a 185 lb person.</span>
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Write one paragraph to explain what circuler motion is?
ZanzabumX [31]

Answer:

When a body moves in a circle with constant speed , it is said to be in uniform circular motion .

Explanation:

  • When an object moves in a circular path , its direction changes at each point .
  • This change in direction result in change of velocity (velocity is vector quantity which changes if direction of the object change) .However speed do not change (it is scalar quantity , not affected by Direction)
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If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this r
EastWind [94]
M_{Al}=26,98\frac{g}{mol}\\&#10;m=51g\\\\&#10;n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\&#10;M_{NaOH}=39,4\frac{g}{mol}\\&#10;m=84,1g\\\\&#10;n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span>  2mol      :     2mol           :                                  3mol
2,14mol   :     1,89mol      :                                  2,835mol
remains         completely consumed
2,14-1,89=0,25mol


A) Al

B)  

M_{NaOH}=39,4\frac{g}{mol}\\&#10;n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g

C)
n=2,835mol\\&#10;M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g

7 0
3 years ago
Borildls03
NISA [10]

Answer:

Mg2+ and Ca2+

Explanation:

Hard water are water that happen to contain salts of calcium and magnesium ions. They can occur as chlorides, sulphates, bicarbonates etc. Hard water wastes soap as it requires an above normal amount to form lathers.

The metallic ions are;

Mg2+ and Ca2+

6 0
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