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Ghella [55]
3 years ago
10

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 135 lb (61.2 kg ) is 5.74×10−3 mol . What volume of a 0.0640 M KCN solution contains 5.74×10−3 mol of KCN? Express the volume to three significant figures and include the appropriate units.
Chemistry
1 answer:
kakasveta [241]3 years ago
6 0

Answer:

The volume will be 89.6875 ml

Explanation:

So to count this we will use a single proportion.

0.0640 mol - 1000 ml

5.74×10−3 mol - x ml

x ml=5.74×10−3 mol*1000 ml/0.0640 mol=89.6875 ml

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Mole ratios how can the coefficients in a chemical equation be interpreted
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Answer:

The coefficient in a balanced chemical equation  indicates the mole ratio of both reactants and products.  

Explanation:

For example lets consider the reation between Hydrogen and Oxygen to form water:

2H2 + O2 ----------------------- 2H2O

In this reaction, the coefficients of the balanced reaction can be transformed to Mole ratio according to Avogadro's Law which states that at standard temperature and pressure, equal volume of gases contain the same  number of moles.

So the mole ratio for the above equation is  the  ratio of the coefficient:

2moles     :     1 mole               :          2 moles

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

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